Reversing a list using two queues
If N is a power of two, I believe O(N log N) operations suffice, even for a somewhat more restricted problem in which all items start on one of the queues and must end up in reverse order on one of the queues (without the input and output lists).

In O(N) steps it is possible to start with all elements on one queue, play "one for you one for me" to split them into alternating subsets on the other queue, and then concatenate them all back into one queue. In terms of the binary representations of the positions of the items, this implements a rotate operation.

In O(N) steps it is also possible to pull off pairs of elements from one queue, swap them, then put them back, reversing all pairs. In terms of the binary representations of the positions of the items, this complements the low order bit of the position.

By repeating O(log N) times an unshuffle and a pairwise swap, we can complement all the bits of the binary representations of the positions — which is the same thing as reversing the list.