It is impossible in general. The reason is that LC Riemannian connection on a compact manifold is automatically complete. This is not the case in the Lorentzian setting.
Edit: Another way to think about this problem is to look at the holonomy group of the connection. For a LC connection holonomy on an $n$-manifold $M$ is always relatively compact in $GL(n,R)$. See here for the complete list of closures of holonomy groups for LC connections on simply-connected manifolds. On the other hand, in the Lorentzian setting, holonomy is contained (up to conjugation) in $O(n-1,1)$ (I am assuming that your pseudo-Riemannian metric is of signature $(n-1,1)$). Thus, the necessary condition is for the holonomy group to be conjugate into a compact subgroup of $O(n-1,1)$.
To be more specific: Suppose that $n=2$ and $M$ is simply-connected. Then the holonomy has to be contained in a connected compact subgroup of $O(1,1)$, which means it is trivial. Hence, your pseudo-Riemannian metric is locally flat. I am pretty sure that this condition is also sufficient for existence of a Riemannian metric with the same LC connection (in the simply-connected case).
Edit: One also has the following. Suppose that $M$ is a simply-connected manifold with (torsion-free) compatible connection $\nabla$, $g$ is a flat pseudo-Riemannian metric on $M$ (signature and dimension do not matter). Then $M$ admits a flat Riemannian metric with LC connection $\nabla$. The same is also true if we replace flatness assumption with the assumption that the holonomy of $\nabla$ is relatively compact (and drop simple connectivity).