actually, scratch that. You are right, it's always true. Let $\rho$ be an arbitrary bipartite state, and $\{\mu(a)\}$ an arbitrary POVM for Alice. The probability of getting the outcome $a$ is $p(a)= {\rm Tr}((\mu(a)\otimes I)\rho)$. The post-measurement result for Bob is $\rho(B|A=a)=\frac{{\rm Tr}_1((\mu(a)\otimes I)\rho)}{{\rm Tr}((\mu(a)\otimes I)\rho)}$. Thus Bob's state knowing Alice performed the POVM $\mu$, but not knowing the outcome, is
$$\sum_a p(a) \rho(B|A=a) = \sum_a {\rm Tr}_1((\mu(a)\otimes I)\rho) = \rho_B$$