Here is my "proof" (from the private conversation mentioned by @renormalize), that if ##A## is hermitian, then ##B## cannot have definite hermiticity...
Suppose ##A^\dagger = A##. Then, $$B ~=~ [A,B]~, ~~~\Rightarrow~~ B^\dagger ~=~
[B^\dagger, A^\dagger] ~=~ [B^\dagger, A] ~=~ -[A,B^\dagger]...