Yes. Let $J$ be the range of $H$ - that is, the class $\{\alpha\in ON: \exists x(H(x)=\alpha)\}$. We can take the Mostowski collapse of $J$, ordered as usual; that is, there is a map $F: J\rightarrow ON$ such that $f(\alpha)=\{f(\beta): \beta\in J\cap \alpha\}$ for all $\alpha\in J$. (This class function $f$ is defined by transfinite recursion.)
Then the map $G=F\circ H$ has the desired property.
The point is that while $R$ may not be set-like, $H$ maps $R$ onto a relation which is set-like in an appropriate way.
Note that the usual caveats about reasoning about classes inside of ZFC apply: really, we're arguing externally that in any model of ZFC, given any definable $R, H$ with the desired properties, there is a desired $G$ with the desired properties.