The answer is no. Any smooth manifold admits a Riemannian metric using paracompactness and partitions of unity: in short, a convex sum of positive definite symmetric matrices is positive definite symmetric. So any manifold has such a structureand is, in particular, Finsler. But there are topolgical topological obstructions to the existence of global pseudo-Riemmannian metrics of a other prescribed signaturesignatures. http://en.wikipedia.org/wiki/Semi-Riemannian_manifold#Properties_of_pseudo-Riemannian_manifolds
EDIT, 16 July: I was looking for an example of said topological obstructions, and with an assist fom Willie Wong it has worked out: the ordinary sphere $$ \mathbb S^2 \subseteq \mathbb R^3 $$ does not possess a signature $(+,-) $ metric which I suppose ought to be called Lorentzian for this dimension. The topological obstruction is that $ \mathbb S^2$ cannot have a smooth (tangent) "line field," just as it cannot have a smooth nonzero tangent vector field by Brouwer. Now, a Lorentzian metric would give (pointwise) null cones, in this case a pair of distinct but intersecting lines in each tangent plane. As we are using $\mathbb R^3, $ we can cheat and define a line field from the angle bisector of the $+$ part of the cone.