Let $(M,g)$ be a Lorentzian manifold and let $R$ be the curvature tensor. We say $R\leq 0$ if 
$$ g(R(X,Y)Y,X) \leq 0\quad \forall \, X,Y \in TM.$$


My question is whether given a Lorentzian manifold $(M,g)$, it is always possible to find a metric $\hat{g}=cg$ such that the curvature of $(M,\hat{g})$ is non-positive.



conformal changes to Lorentzian curvature
No, this is not possible if the Lorentzian manifold has a null geodesic admitting a pair of conjugate points.
The notion of conjugate points for null geodesics does not depend on the conformal factor.
By the Lorentzian version of the Cartan-Hadamard theorem (see e.g. the book by Beem et al. 1996 Global Lorentzian Geometry. Prop. 11.13) in a spacetime if the inequality you wrote were to hold for any pair of timelike vectors then there would not be causal geodesics with conjugate points.


