51min10sec
1. F
2. (B2)
3. (U' F2 L')
4. (B2 U R')
5. R2
6. U2
7. (R B2 L' B L B' R' B') {3e/3'1'c}@[18]
Alt.
7. (B L' B2 L R B' R') {4e/4'1'c}@[17]
8. (B) {3e/5c}@[18]
Alt.
7. (B2 L' B L)
8. (R B' R' B) {3e/5c}@[18]

Went with last {3e/5c} option, in normal skeleton found:
2-1: U' [R' D R D'] U -> {3c} +[6]
8-3: B L B' R2 B L' B' R2 -> {} +[6]

Tempting to see if I may have gotten lucky with {4e/4'1'c}@[17], say net [5]ish -> {3c}, and I had a little extra time where I could have tried to check this out quickly, but being pressed for time, I probably would have messed something up anyway.

This maybe had some potential, but a lot of option to have to explore, and I just messed up and ran out of time in the end.
{22e/2'2'c} [20]
Found
net [1] -> {3e/5c}
net [5] -> {5c}
net [0] -> {3e/5c} then a net [4] -> {5c} but in the interest of time, I should have just taken the {5c}@[25], and there are still a lot of things to look for in a {22e/2'2'c case} that I just don't have the ability to focus on them all simultaneously, so it takes forever.

1. L2 D B2
2. R2 F' L
3. B D B2
4. D' R
5. (R2 D2 R) {3'1'c/5c}@[11]

1. L2 D B2
2. R2 F' L
3. B D B2
4.
5. D R2
5-2: R' U L [D2 F D2 F'] L U R {3e/3c}@[20]
or
3-1: R' [F D F' D'] R {3e/5c}@[17], went for this over the prior one

1-0: R2 U2 [D2 B D2 B'] U2 R2 -> {3c}@[25]
1-0-4: B D' F' D B' D' F D -> {}@[31]

Just got this corrected with 1 second left!
Not bad for kind of a disaster of a solution.
{3e/2'2'c} @[17] couldn't really find anything, but I forget how flexible this configuration can be because I remain very fixated on solving the unsolved corners, but it is also very appropriate for my insertion to only involve say 3 of the 4 corners and just create an extra unsolved corner resulting in a {5c} case, or maybe just solving one corner but adding an unsolved one as I found a net [5] to {3'1'c} and was then able to find +[5]+[5] -> {}. If I can remember to consider these options, then this may more easily result in some < net [4] move insertions to solve the edges without making the corner situation worse.
1. R
2. B' L' F2
3. R' D2
4. F
5. (F)
6. (L B L B')
7. (L' U' L2)
8. (L U2 L')

In normal space skeleton:
7-1: F' [F' D F D'] F -> {3'1'c}
5-1: F L' B' L F' L' B L -> {3c}
8-3: L' U' L D L' U L D'

I initially wrote this last one down with U and U' interchanged, which is the last second correction I was able to make and avoid a DNF.

1. (F' L F2)
2. (U' L2)
3. (R F D2)
4. R U F2 U'
5. R2 F' R
6. F' R2 {3e/2'2'c} @[17]
Alternates:
6. (R2) {4e/21'1'c} @[15]
7. (F') {3e/22c} @[17]
7. (F) {3e/2'2'c) @[17]

Went with first {3e/2'2'c} @[17]
5-1: R F2 F' F2  +[2] -> {5c} +[7]+[6] -> {}
Can't complain about {3e/2'2'c} -> {5c} in net [2], but hoped for better than +[13] on the {5c}. Who knows, if I actually had time to fully check it out, perhaps I would have found something in fewer moves.

{3e/22c}@[21] +[5] -> {2'2'c} +[4]+[7] -> {}

Well, that ended up being okay as a mad dash just to get something after struggling with the skeleton and making all sorts of turning errors.
{3e/5c}@[22] +[9] -> {3c} +[4] -> {}

Initially was going to go with a {3e/5c} +[7] -> {5c}@[29], but took my chances with a few extra moments to look for something better.

1. F2 U L'
2. R2 U2
3. F2 L2 F2
4. L
5. (L D2 L)
6. (F' D2 F)
7. (D) {3e/31'1'1c} @[16]
Alt.
7. (D') {3e/5c} @[16]
Alt.
6. (D' F' D' F D') {3e/2'2'2c} @[17]

Went with {3e/2'2'c}, @4-0 found: D2 B R2 U [D' L D L'] U' R2 B' D2 -> {} @[29]
Thought I found a way to do the same nearby with net [10] instead of net [12] but must have messed something up because I couldn't reproduce it. I was hoping to have some time to explore {3e/5c}@[16] case, but ended up wasting a lot of time dealing with trying to reproduce a [27] that I don't think existed, and then it was scary for a couple minutes because I even messed up reproducing the [29] I thought I had secured, but that was realtively easy to resolve since I think I just made a misturn and was pretty confident it worked because I checked it earlier.

I'm also kind of stupid because I should have just looked for a shorter way to just get to {3c}, and in going back, sure enough I found:
3-2: L' [D' R' D R] L; i.e., +[5] {3e/2'2'c} -> {3c} [22], and then within this I found
3-2-3: L2 D R D' L2 D R' D'; i.e., +[4] -> {}

Not even sure that's the best option, but it's pretty decent.

1. F2 U L'
2. R2 U2
3. F2 L2 F2
4. L
5. (L D2 L)
6. (F' D2 F)
7. (D) {3e/31'1'1c} @[16]
Alt.
7. (D') {3e/5c} @[16]
Alt.
6. (D' F' D' F D') {3e/2'2'2c} @[17]

Went with {3e/2'2'c}, @4-0 found: D2 B R2 U [D' L D L'] U' R2 B' D2 -> {} @[29]
Thought I found a way to do the same nearby with net [10] instead of net [12] but must have messed something up because I couldn't reproduce it. I was hoping to have some time to explore {3e/5c}@[16] case, but ended up wasting a lot of time dealing with trying to reproduce a [27] that I don't think existed, and then it was scary for a couple minutes because I even messed up reproducing the [29] I thought I had secured, but that was realtively easy to resolve since I think I just made a misturn and was pretty confident it worked because I checked it earlier.

{3e/5c}@[19] + [10] -> {3c} + [6] -> {}

This solution was quite fun!
1. R F2 D'
2. U' F2 R' {+1 to 2x2x3 with switch to I-space}
3. R' L F'
4. L' F L' R2 U2 R' {+1 to F2L with switch to I-space}
5. (L U) {3e/3'1'c} @[16]
6. (L' U' B' U B L) {3c} @[22]
Was pretty happy with seeing {2x2x3} @[7], but noticed I could just keep going and quickly get to F2L-1 and even full F2L. Ending up with {3c}@[22] is already great for me, so I figured I most likely could secure a sub-30, and sure enough found a net [6] to have [28] as a backup solution, and had all of this with about 40 minutes left.

Then I worked the {3e/3'1'c} skeleton and found:
4-5: U [R B' R' B] U' net [5] to {3c}
4-4:  R' [F' U F U'] R net [5] to {3c}

Went with the insertion at 4-4 and found at 3-3: F L B' L' F' L B L' net [4] to solve.
Also went back to check the 4-5 insertion, but it resulted in the exact same 3c cycle as the 4-4 one. I think I found one 3c cycle that was a net [5] somewhere in step 4, but that was irrelevant at this point.

Am curious how things might have turned out if I made the switch to I-space after step 2, but I didn't have time for this?
Well I played with it some and got {3e/2'2'1'c}@[16], which isn't all that different from {3e/3'1'c}. There's also {3'2'e/4'1'c}@[11], which would probably be a little much for me to deal with in one hour.

Well since I've already submitted a timed solution:
1. R F2 D'
2. U' F2 R'
3. (L' F' L' U2 R U R') {3'2'e/4'1'c}@[11]
4. (U' F' U) {3e/2'2'1'c}@[16]

Most interesting things that came out of:
1. R F2 D'
2. U' F2
3. U' R' U2 L F L

3-4: U' L' U [F'] U' L U {3e/5c}@[18]
3-1: U' F [U'] F' U {3e/2'21'c}@[15]
2-1: F' U R [U] R' U' F {3e/3'1'c}@[17]

Worked through {3e/2'21'c}@[15] case, but don't really see anything getting me better than [25]

Had an ugly {3e/3'1'c}@[25], but was able to work it a little differently and get the same @[20] at least.

Then found two places where +[10] -> {3c}, but only had time to check one of them out. Didn't help that I made a misturn while looking for the final insertion, but I was a bit lucky and quickly found a net 5-mover to solve after solving and re-scrambling, whereas I thought I only had a net 7-mover before that.

{3e/2'2'1'c}@[20] vs. {3e/31'1'c}@[20] vs. {4e/2'3'c}@[19]

Took: {3e/2'2'1'c} +[8] -> {3c} +[7] -> {}

I suppose there's a chance the {4e/2'3'c} could end up better, but I only have 5-6 minutes left.

Didn't get great skeletons: {3e/1'1'c} vs. {3e/22'1'c}, both @[21]

{3e/22'1'c} +[4] -> {5c} +[5]+[5] -> {}

Also had {3'1'c}@[26], but didn't have time to completely examine {5c} case, which was pretty sparse for about the last half of the skeleton; i.e., all 7 or 8 movers, if anything at all.

Don't care!

{4e/2'1'c} @[19] vs. {4e/2'1'1'c} @[21]

Went for first one, but couldn't find anything and forced +[11] -> {2'2'c} but was pretty much out of time at that point, and now I just made two crappy 8-move 3c cycles to finish.

{3e/2'2'1'c} [22]
{4e/2'3'c} [20]
{3e/31'1'c} [21]
{5c} [27]

Didn't really have time to explore anything, wanted to at least try the {3e/31'1'c} case, but I only had a little less than 10 minutes left, and it isn't a very good case for me anyway. So I just rushed out two 3c-cycles to finish the {5c} case. I didn't even reconstruct the {5c} skeleton in the normal space and just worked within the moves of the last normal session, but given the circumstances, finding {5c} +[6]+[6] -> {} isn't too bad.

{3e/5c}@[19] with {5c}@[25] backup

{3e/5c}@[19] +[4] -> {5c} +[6]+[5] -> {}

{2'1'e/4c}@[20] + [4] -> {3e/3c} + [7] -> {3c} + [7] -> {}
Well at least I didn't DNF.

This was cool...I tried something I never really would have done before!
1. (B')
2. B L2
3. R' D2 R' D
4. F U F' U'
5. F D' F' D {4e/4c} [15]
6. D R2 D' R2 {2e/2c} [18]
Alt 6. D R D' R' {2'1'c/2'1'e} [18]
Alt 5. F' R2 F {4e/4'1'c} [14]

I went for {4e/4c} option and found several insertions that weren't terrible: {5c}@[22], {3e}@[30], {3c}@[28], {3e}@[26], but what was interesting is that at 4-2, a simple [F] move, which is net [-4] got me {22e}@[11]; i.e., it solves all the corners and one edge, but also introduces an unsolved edge. I figured, what the heck, let's look for two 3-edge cycles.

So now my skeleton is:
2. B L2
3. R' D2 R' D
4. F2
5. D' F' D
1. B

At 4-1: F [F2 D U' R2 D' U] F' for {3e}@[17]
I didn't have time to think carefully about the second insertion and just put
U [L2 F' D' U L2 D U' F' L2] U' at the end for {3e} + [11] -> {}. I did search through the entire skeleton for the 2nd insertion, but was a bit rushed. There were also other variations I may have been able to use for the first one, and I didn't look through steps 2 & 3 for the first insertion because I felt the net [6] was pretty good for a 3-edge cycle.

Anyway, I generally try to avoid pure edge cycles, but I figured with 4 edges, it's likely three of them would end up in a decent configuration for the first one, especially since it was {22c}; i.e., doesn't matter which way the edges are sent so regular and inverse algs can be considered. Then it was really about finding the last one.

Did this all on the inverse scramble, so in normal space, my skeleton is as follows.
5. U F U F'
4. R' F2 R F2 U F2
3. L' F2 L2 U'
2. D2 B2
1. L2 {3e/3'1'c} [17]

Insertions that fix edges and perhaps corners
3-3: L' [F U F' U'] L {4c-type, most likeley 3'1'c} [22]
3-3: L' F D [U F' U' F] D F' L {3c} [26]
4-5: [F' U F U'] {3'1'c} [21] went with this

2-2: B2 D' F' D B2 D' F D -> {3c} [26]
2-0: [D2 B2 D' F' D B2 D' F D']' -> {3c} [26]

Went with 2-2 3c-cycle and found @3-2: F2 L' B L F2 L' B' L net [4] -> {}

1. R2
2. (D)
3. (R' B)
4. B'
5. (L' U2 F' U' F2 U')
6. D' L' D
7. D F' D'
8. (F2) {3e/3'1'c} @[17]
9. (B D L D' L' B') {3c} @[23]
Went for {3e/3'1'c} @[17], but only decent insertion I found was equivalent to step 9. Found a net [3] for {3c}->{} at 5-2 in normal skeleton: F2 U B' U' F2 U B U'

Stupid lately. I'll always write some insertion off to the side on the paper, and then forget to include it when making the turns to look for another insertion :(

Out of time :(

1. F' R F2
2. L B2
3. L' B L2 D F' D' {4'1'e/3'2'c} [11]

3-4: L D L U' [R'] U L' D' L' {3e/22'c} [20]
or
1-2: U L' [U'] L U' {3e/5c} [16]...went with this and found 3-0: D2 [R F U F' U' R'] D2 -> {5c} [24]
found a quick +[5], but only a +[8] was compatible for the 2nd 3c-cycle.

Out of time :(

{4e/4'1'c} @ [16] + [8] -> {2'2'c} +[6]+[6] -> {}

{3e/5c} [21] +[5] -> {5c} +[7]+[5] -> {}

1. D R2 D' F D
2. U' F' U2
3. F' L2 F2 L'
4. (D' L' F2) {3e/5c} {15}

@1-5: U2 L R' F2 L' R {3e/5c} -> {5c} (pure 3e cycle)
+[5] +[5] {5c} -> {}

{3'1'e/22c} [19] vs. {3e/5c} [20] vs. {3e/22'1'c} [17]

Went for {3e/22'1'c} [17]
+[5] -> {3'1'c} +[5]+[7] -> {}