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No. Suppose otherwise. Let $G$ be the Levy collapse generic, and $D$ be the generic for forcing with the nonstationary ideal after that. Since $\mathrm{Ult}(L[U,G],D)$ is wellfounded, letting $i_D:L[U,G]\to\mathrm{Ult}(L[U,G],D)$ be the ultrapower map, we have that $i_D(L[U])=L[i_D(U)]$ is some iterate of $L[U]$. Let $j:L[U]\to L[i_D(U)]$ be the iteration map. Let $\Gamma$ be a proper class of ordinals fixed by both $j$ and $i_D$. We have $j(U)=i_D(U)$ and $j(\alpha)=i_D(\alpha)$ for all $\alpha\in\Gamma$. But $L[U]=\mathrm{Hull}^{L[U]}_{\Sigma_1}(\Gamma)$ (using that $\Gamma$ is proper class and definable in a generic extension of $L[U]$). It follows that $j=i_D\upharpoonright L[U]$. But working back in $L[U]$, because this Levy collapse is countably closed, the set $X$ of all $\alpha<\kappa$ of cofinality $\omega$ remains stationary in $V[G]=L[U,G]$. So we could have taken $X\in D$, giving that $\kappa\in i_D(X)$, whereas $\kappa\notin j(X)$, but $X\in L[U]$, a contradiction.