Is the existence of an oracle such that $P^O = NP^O$ nontrivial?
The existence of an oracle with respect to which $\mathsf{P} = \mathsf{NP}$ shows that $\mathsf{P} \neq \mathsf{NP}$ cannot be proved using techniques that relativize such as diagonalization. In contrast, the time and space hierarchy theorems do relativize, disproving your claim that any two complexity classes are equal relative to some oracle.

The easy observation that $\mathsf{P}^\mathsf{R} = \mathsf{EXP}^\mathsf{R}$ seems to contradict the fact that the time hierarchy theorem relativizes. Tthe definition of $\mathsf{X}^\mathsf{R}$ that you use is $$ \mathsf{X}^\mathsf{R} = \bigcup_{A \in \mathsf{R}} \mathsf{X}^A.$$
The time hierarchy theorem shows that $\mathsf{P}^A \neq \mathsf{EXP}^A$ for every language $A$, but not for every family of languages.

The oracle with respect to which $\mathsf{P} = \mathsf{NP}$ is a single language. One possible choice is any $\mathsf{PSPACE}$-complete language. This is very similar to using an arbitrary recursive language, with a crucial difference – there are languages in $\mathsf{PSPACE}$ which are complete with respect to polynomial time reductions.