The quantum entropy is concave. When does equality exactly hold $$H\bigg(\sum_x p_X(x) \rho^x\bigg) = \sum_x p_X(x) H(\rho^x)$$
It holds, for example, if $\rho^x$ is a fixed state independent of $x$.
Is there another case? Is the entropy of a mixture of pure states also zero?
As per Nielsen and Chuang's textbook in its 10th Anniversary Edition, equation $(11.84)$:
$$\sum_ip_iS\left(\rho_i\right)\leqslant S\left(\sum_ip_i\rho_i\right),$$ which is concavity! Note that equality holds if and only if all the states $\rho_i$ for which $p_i>0$ are identical; that is, the entropy is a strictly concave function of its inputs.
Thus, there is no other case. You can find the proof in the aforementioned textbook, which I'll reproduce here.
We first introduce a $B$ register to define a bipartite state $\rho^{AB}$: $$\rho^{AB}\overset{\text{def}}{=}\sum_xp_x\rho_x\otimes|x\rangle\!\langle x|\,.$$ The entropy we're interested about is $S\left(\rho^A\right)$, which is equal to: $$S\left(\rho^A\right)=S\left(\sum_xp_x\rho_x\right)\,.$$ We can also compute the entropy of the $B$ system: $$S\left(\rho^B\right)=S\left(\sum_xp_x|x\rangle\!\langle x|\right)=H(p)$$ with $H$ being the Shannon entropy. Finally, the joint entropy is given by: $$S(A, B)=H(p)+\sum_xp_xS\left(\rho_x\right)$$ by the Joint entropy theorem (Equation $(11.58)$ in the textbook). But now, the subaddivity property of the joint entropy states that: $$S(A, B)\leqslant S(A)+S(B)$$ which in our case gives: $$H(p)+\sum_xp_xS\left(\rho_x\right)\leqslant H(p)+S\left(\sum_xp_x\rho_x\right)\,.$$ Which is the inequality we seeked. Thus, the question becomes: when is the subadditivity inequality an equality? We can find the answer on Equations $(11.74-11.76)$: we have equality if and only if $\rho^{AB}$ is a product state (Theorem $11.7$), that is if $\rho^{AB}=\rho^A\otimes\rho^B$. But now, note that we have: $$\rho^A\otimes\rho^B=\sum_{x, y}p_xp_y\rho_x\otimes|y\rangle\!\langle y|\,.$$ In particular, we must have for all $x$: $$\left(I_A\otimes\langle x|\right)\rho^{AB}\left(I_A\otimes|x\rangle\right)=\left(I_A\otimes\langle x|\right)\left(\rho^{A}\otimes\rho^B\right)\left(I_A\otimes|x\rangle\right)$$ which gives us: $$p_x\rho_x=p_x\sum_yp_y\rho_y$$ which gives us, assuming $p_x>0$: $$\rho_x=\sum_yp_y\rho_y\,.$$ In particular, $\rho_x$ does not depend on $x$, which concludes the proof.
