(Not a full answer, but too long for a comment)

**Fact:** there are regular graphs where every equitable coloring has color classes with no edges between them. (As mentioned in comments, if we remove the restriction that the coloring be equitable, the question is no longer interesting.)

Moreover, there are regular, perfect graphs satisfying the above as well (so be careful).

Let $H$ be any $r$-regular graph on $m$ vertices with $\chi(H) > m / (m-r-1)$. Then construct an $m$-regular graph by adding an independent set $S$ of $m-r$ new vertices, which are each connected to all the vertices of $H$. Any coloring of the type we want must have $S$ as one of its independent sets. This means there's a coloring like you want iff there is an equitable coloring of $H$ (of the form you like) having color classes of size roughly $|S| = m-r$. But since $\chi(H) > m / (m-r-1)$, the graph $H$ has no coloring where the sets all have size at least $m-r-1$. Thus, the graph constructed in this way has no coloring of the type we're looking for.

By starting with various graphs $H$, this can be used to construct lots of regular graphs without any colors that you want. For instance, if $H$ is perfect, this construction gives a perfect graph without these colorings.

For concreteness, if $H$ is the disjoint union of $2$ triangles, then $\chi(H) = 3 > m/(m-r-1) = 6 / (6-2-1) = 6/3 = 2$. And the above construction gives a graph with $10$ vertices and no equitable coloring of the type you're asking for.