Oops, what I originally wrote assumed U=X; didn't read the question properly. Here's a fix:
First, suppose $L$ is simple. $L$ is trivialied by some finite etale map $\tilde U\to U$ (this is what it means to have finite monodromy); then applying Grauert-Remmert (or SGA I, as mentioned in Georges Elencwajg's answer to this question), we may extend the map to a (possibly ramified) map $\pi: \tilde X\to X$. That is, $\pi^*L|_{\tilde U}=\underline{\mathbb{C}}^n$. Then $\operatorname{Hom}|_U(L, \pi_*(\underline{\mathbb{C}}^n)|_{U})\simeq\operatorname{Hom}_{\tilde U}(\pi^*L|_{\tilde U}, \underline{\mathbb{C}}^n)$ is non-zero, so $L$ is a factor of $\pi_*\underline{\mathbb{C}}^n$ (using simplicity). Again, as $L$ is simple, it is in fact a factor of $\pi_*\underline{\mathbb{C}}$.
Now if $L$ is not simple, decompose it into simple factors $L=\oplus L_i$, and let $\pi: {\tilde X_i}\to X$ be covers trivializing the $L_i$. Setting $\tilde X=\bigsqcup {\tilde X_i}$ does the trick.