For a random oracle R, does BPP equal the set of computable languages in P^R?
Yes.

First, since it took me a minute to figure this out myself, let me formalize the difference between your question and $\mathsf{AlmostP}$; it's the order of quantifiers. $\mathsf{AlmostP} := \{L : Pr_R(L \in \mathsf{P}^R) = 1\}$, and the result you allude to is $\forall L\, L \in \mathsf{BPP} \iff Pr_R(L \in \mathsf{P}^R) = 1$. If I've understood correctly, you are asking if $Pr_R(\forall L\, L \in \mathsf{P}^R \cap \mathsf{COMP} \iff L \in \mathsf{BPP}) = Pr_R(\mathsf{P}^R \cap \mathsf{COMP} = \mathsf{BPP}) = 1$.

Consider 

$p := 1-Pr_R(\mathsf{P}^R\cap \mathsf{COMP} = \mathsf{BPP}) = Pr_R(\exists L \in \mathsf{P}^R \cap \mathsf{COMP} \backslash \mathsf{BPP})$. 

By the union bound, the $p$ is upper-bounded by $\sum_{L \in \mathsf{COMP}} Pr_R(L \in \mathsf{P}^R \backslash \mathsf{BPP})$. (Note that the latter sum is countable.) Now, by the 0-1 law - which applies since all the relevant statements do not change if we change $R$ finitely much - each individual probability in this sum is either 0 or 1. If the answer to your question is no, then $p=1$, so there must be some $L \in \mathsf{COMP}$ such that $Pr_R(L \in \mathsf{P}^R \backslash \mathsf{BPP}) = 1$. But this contradicts the fact that $\mathsf{AlmostP} = \mathsf{BPP}$.

Update Oct 10, 2014: As pointed out in the comment by Emil Jeřábek, the same argument applies to $\mathsf{AM}$ vs. $\mathsf{NP}^R$, since we also know that $\mathsf{AlmostNP} = \mathsf{AM}$. 

He also points out that we didn't use anything about $\mathsf{COMP}$ other than that it is a countable class that contains $\mathsf{BPP}$ (resp., $\mathsf{AM}$). So the "interesting conclusion" in the OQ actually applies to any countable class of languages $\mathcal{C}$ that contains $\mathsf{AM}$: if $\mathsf{P} = \mathsf{NP}$, the "only" languages that witness the oracle separation $\mathsf{P}^R \neq \mathsf{NP}^R$ are outside of $\mathcal{C}$. But the latter statement feels somewhat misleading to me (it makes it sound like, for any $L_0$ we could consider $\mathcal{C} = \mathsf{AM} \cup \{L_0\}$, and thereby "show" that no $L_0$ realizes $\mathsf{NP}^R \neq \mathsf{P}^R$, contradicting the well-known theorem). Rather, writing it out symbolically, we've shown:

If $\mathsf{P} = \mathsf{NP}$, then $\forall \text{countable } \mathcal{C} \supseteq \mathsf{AM}\, Pr_R(\mathsf{NP}^R \neq \mathsf{P}^R \text{ and } \mathsf{NP}^R \cap \mathcal{C} = \mathsf{P}^R \cap \mathcal{C}) = 1$.

Note that, crucially, probability 1 is not the same thing as all $R$, and which full-measure set of $R$ satisfy the argument to $Pr_R$ can depend on $\mathcal{C}$. So if we try to alter $\mathcal{C}$ to $\mathcal{C} \cup \{L_0\}$, it at most removes a measure 0 set of $R$ that satisfy this statement.