I think the answer is no. If $\kappa$ is huge, and we derive an almost-huge tower from the huge embedding, then the embedding computed from the tower has stationary correctness.
Suppose $h : V \to M$ is a huge embedding and $h(\kappa) = \lambda$. Then $M$ computes stationary subsets of $\lambda$ correctly by the closure of $M$. As in Kanamori's proof, let $\langle U_\alpha : \kappa \leq \alpha < \lambda \rangle$ be the almost-huge tower derived from $h$. So $U_\alpha = \{ X \in [\alpha]^{<\kappa} : h[\alpha] \in h(X) \}$. The argument in Kanamori's book shows that this is a tower giving rise to an almost-huge embedding with the same target for $\kappa$.
For each $\alpha < \lambda$, we have the ultrapower embedding $j_\alpha : V \to M_\alpha$, and we have factor maps into the direct limit $k_\alpha : M_\alpha \to M_\infty = N$. The critical point of each $k_\alpha$ is greater than $\alpha$. Let $j : V \to N$ be the limit embedding.
I claim that for any $X \subseteq \kappa$, $j(X) = h(X)$, so stationary sets are mapped to stationary sets. Of course $j(X) \cap \kappa = X$, so let $\kappa \leq \alpha < \lambda$. We have:
$$\alpha \in j(X) \Leftrightarrow \alpha \in k_\alpha \circ j_\alpha(X)$$
$$\Leftrightarrow \alpha \in j_\alpha(X) \text{ (since crit}(k_\alpha) > \alpha)$$
$$\Leftrightarrow \{ z : \mathrm{ot}(z) \in X \} \in U_\alpha$$
$$\Leftrightarrow \mathrm{ot}(h[\alpha]) \in h(X)$$
$$\Leftrightarrow \alpha \in h(X).$$
We don't need the full strength of hugeness to get an almost-huge tower with this property, just because the statement "there is an almost huge tower with stationary correctness" will reflect below a huge cardinal. (But I suppose we wouldn't think it's consistent without proving it from a simpler principle like hugeness.)