Deforming metrics from non-negative to positive Ricci curvature

There are obstructions. Perhaps the most famous comes from the theorem that, if a compact spin manifold has a metric of positive scalar curvature, then its $\hat A$-genus must vanish.

If you take a compact Riemannian spin manifold $(M,g)$ with special holonomy $\mathrm{G}_2$ (in dimension $7$), $\mathrm{Spin}(7)$ (in dimension $8$), or holonomy in $\mathrm{SU}(n)$ (in dimension $2n$) whose $\hat A$-genus is nonzero (and there are lots of these, even simply-connected ones), then $g$ will be Ricci-flat and hence will have non-negative Ricci curvature. However, by the above theorem, it cannot carry any metric with positive scalar curvature, let alone a metric with positive Ricci curvature.

This is not a complete answer but would be helpful. Here are a few facts:

Theorem (T. Aubin 1979). If the Ricci curvature of a compact Riemannian manifold is non-negative and positive somewhere, then the manifold carries a metric with positive Ricci curvature.

Also vanishing the first Betti number is a necessary condition in compact case for admitting strictly positive Ricci curvature (see on Google books: A Course in Differential Geometry, By Thierry Aubin).

Relation with scalar curvature:

1. There are still no known examples of simply connected manifolds that admit positive scalar curvature but not positive Ricci curvature

2. If a manifold $M$ cannot have a metric with positive (or zero) Scalar curvature, then it certainly does not admit a metric with positive (zero res.) Ricci curvature.

This paper of G. Perelman is also useful: "Construction of manifolds of positive Ricci curvature with big volume and large Betti numbers". Comparison Geometry. 30: 157–163 Click here for pdf