As Andreas suggests, I shall fix $\mathcal U$ to be countably-incomplete. In fact, wlog, $\mathcal U$ will be over $\mathbb N$. If $X$ is not super-reflexive, then you don't even get all the rank-one operators. We know that $(X)_{\mathcal U}^* = (X^*)_{\mathcal U}$ if and only if $X$ is super-reflexive, so there is $\lambda \in (X)_{\mathcal U}^* \setminus (X^*)_{\mathcal U}$. Choose $y=(y_n)\in (X)_{\mathcal U}$. Let $T(x) = \lambda(x)y$ so $T$ is a rank-one map on $(X)_{\mathcal U}$. Suppose $T=(T_n)$. For each $n$ pick $\mu_n\in X^*$ with $\|\mu_n\|\leq 1$ and with $\lim_n \mu_n(y_n)=\lim_n \|y_n\|$ (limits over $\mathcal U$ of course). Set $\mu=(\mu_n)$. Then
$$ \mu(T(x)) = \lambda(x) \mu(y) = \lambda(x) = \mu((T_n)(x)) = \lim_n \mu_n(T_n(x_n)), $$
which holds for all $x$, so $\lambda = (\mu_n\circ T_n)\in (X^*)_{\mathcal U}$, contradiction.

If $X$ is super-reflexive, then I want to use some "co-ordinate" structure, so I need to think some more...