The answer is YES. There are some "Pseudo $3\times 3\times 3$" $4\times 4\times 4$ positions which, although they can be exclusively solved by turning outer layer faces only (e.g., using a $3\times 3\times 3$ solver), if one applies certain $4\times 4\times 4$ move sequences (which consist of both inner and outer slices), the resulting solution will be fewer moves than optimal solutions retrieved from a $3\times 3\times 3$ solver.
This is a "trivial position" found by Stefan Pochmann in the speedsolving.com thread, Pseudo $3\times 3$ shorter than $3\times 3$?.
As you can see, I was a participant in that thread as well, and I found the following more complicated position. (I was motivated to find what may be considered "non-trivial" positions of this type -- to show that Stefan's wasn't the only $4\times 4\times 4$ position like this.)
Uw2 2R2 Uw2 Rw2 U2 Rw' D' U' R2 D' U' s2 Rw (15 htm, 13 stm)
z2 x' B2 L U D R2 B e' m' B' U' D' L' s' U2 (17 htm, 14 stm)
(You may verify that 17 htm is move optimal from any move optimal 3x3x3 computer solver you wish.)
In fact, based on the eight short PLL parity fixes that I listed in the last post of that thread (I eventually found four more and put all $12$ in this section of the $4\times 4\times 4$ parity algorithms speedsolving wiki page), we can make dozens of such examples by combining two short PLL parity move sequences together.
I never explored the topic further to know if there is another method for finding such positions; and therefore, I cannot claim that my method is the only method for finding them.