Sort of cheating, but you could do this:
[R2 U F2 U' F2, {mirror across x=z plane}]
= R2 U F2 U' F2 R2 U' R2 U F2
R2 U F2 U' F2 causes a bowtie-shaped 4-cycle of edges between the U and D layers. This cycle shape is preserved under the mirroring, so applying the inverse after the mirroring causes the U- and D-layer edges to be solved again. It also sends the DFL, DFR, DBR corners to the U layer in a symmetrical manner (so again, the commutator doesn't affect those corners), but it sends the UFL, UFR, UBR corners to the D layer in an asymmetrical manner, which ends up being an A-perm at the end of the commutator.
R2 U # F2 U' F2 R2 U' R2 U F2
Insert [E', F2] at # to get rid of the 3-cycle of edges in the E slice and simplify to get R2 D R2 D' F2 R2 U' R2 U F2.