My question is:
Can we judge a manifold that can admit a (p,q) metric?
I only know the case that the existen of a lorentz metric is equivalent to Euler Character is zero
The criterion for existence of a $(p,q)$ metric is (assuming $p+q=dim X$) that the tangent bundle splits as a direct sum of two subbundles of dimensions $p$ and $q$.
EDIT : I doubt there is an easy algebraic topology criterion in general, as characteristic classes [EDIT after Lennart Meier's comment: other than Euler class, which cannot help if $p,q>1$] are stable invariants of vector bundles. At least, one has necessary conditions, such as : the total Stiefel-Whitney and Ponttryagin (Chern of complexified bundle) class of $TX$ is the product of two (inhomogeneous) classes of degrees at most $p$ and $q$. Perhaps real $K$-theory (and its operations) gives better necessary conditions (although still not sufficient in general). But it's not easy to compute.
I believe that you're a bit mistaken about the final claim. The correct statement should be that:
The proof of this is easy:
(the only if direction is trivial) Suppose there is a nowhere vanishing vector field $X$. Pick any Riemannian metric $g$ and let $\omega$ be the dual $1$-form with $X$ with respect to $g$. Then, defining $$ \tilde g = - 2\omega\cdot\omega + g, $$ this is a "time orientable lorentz metric" as desired.
The same proof shows that
This is something which can be detected by Euler classes, I think.
I agree that this second conclusion is a bit unsatisfying, because it is natural to restrict to time orientable Lorentz metrics for physical reasons, but here it is not clear that it is a natural restriction.