I would personally try it by solving the top and bottom layers like a domino, then the middle 4 layers similar to a 6x6. middles, then edges. the same parity algorithms from the larger x by x by x cubes would work. I am sure there is a more efficient way of doing it, but that should get the job done.
Are you aware that this puzzle has four faces that can be turned by 180 degrees only? In Burgo’s post:
Parity Alg: (Uu)2 F R u2 R F (Uu)2 =(centre fix) the rest is domino algs as Maarten said.
the R and F moves are a shorthand for F2 and R2, because nothing else can be done.
I do not understand how the analogy with 6x6x6 centre pieces helps?
If you can solve a 3x3x4, you can solve this 3x3x6 as well. As pointed out already, you have just 3 Dominos in one puzzle.
The 3x3x4 can be viewed as two Dominos. A parity problem occurs on the 3x3x4 as well.
So, solving a 3x3x6 should be straight forward, if you can solve a 3x3x4. Maybe it is a little bit more confusing.
BTW, if you want to use Burgo's parity algorithm you have to adapt it for a parity on the outer (U/D white/yellow) layers. The current WCA notation does not have a notation for third layer tutns, therefore I introduce 3u for the layer under u (and u is under U. WCA has 3U for turning three layers together.
It becomes: 3U2, F2, R2, (u3u)2, R2, F2, 3U2