As for $triR_k(K_{3,3})$, I would assume that the usual `averaging' approach should give an estimate not far from the optimal one. Take three colors with the maximal number of edges in them and estimate the number of triples of vertices having a common neighbour by these colors. If this number exceeds $2{n\choose 3}$, we are done.

Exactly on the described situation, it seems that you have no chance for finding $K_{3,3}$ even in a complete graph with $n$ vertices and $n$-colored edges. Indeed, if we take a regular $(6k+1)$-gon and paint all parallel edges in one color, then the resulting graph would not contain a trichromatic $K_{3,3}$. Moreover, it seems that here we have some regularity which helps finding such $K_{3,3}$...

As for $triR_k(K_{3,3})$, I would assume that the usual `averaging' approach should give an estimate not far from the optimal one. Take three colors with the maximal number of edges in them and estimate the number of triples of vertices having a common neighbour by these colors. If this number exceeds $2{n\choose 3}$, we are done. [EDIT] This seems to be covered by the Kővári–Sós–Turán theorem, see https://en.wikipedia.org/wiki/Zarankiewicz_problem. [END EDIT]

Exactly on the described situation, it seems that you have no chance for finding $K_{3,3}$ even in a complete graph with $n$ vertices and $n$-colored edges. Indeed, if we take a regular $(6k+1)$-gon and paint all parallel edges in one color, then the resulting graph would not contain a trichromatic $K_{3,3}$. Moreover, it seems that here we have some regularity which helps finding such $K_{3,3}$...