Let $Q$ and $R$ be subgroups of the finite $p$-group $P$ and suppose that $Q$ is normal (maybe this is not needed). Is it then true that $$N_Q(R)=N_Q(Q\cap R)?$$ Obviously, we have $N_Q(R)\le N_Q(Q\cap R)$, but the other direction is unclear to me. For suppose $y\in N_Q(Q\cap R)$ so that $Q\cap yRy^{-1}=Q\cap R$. From here, I do not see how to reach the conclusion $R=yRy^{-1}$ and in fact in general it seems to be false.
If the question is not true, do there exists some condition one can put on $Q$ or $R$ so that it is true?
No, this is not true in general. For example, we might have $Q \cap R = 1$ yet $Q$ does not normalize $R$.
For an example, consider a nontrivial semidirect product $G = R \ltimes Q$ (of two $p$-groups in your assumptions, but this is not really relevant here). Then $Q$ is normal, $N_Q(R \cap Q) = N_Q(1) = Q$. But $N_Q(R)$ is not all of $Q$, because otherwise $R$ would be normal. One particular example like this is given by the dihedral group of order $8$.
I don't know if there are any interesting conditions that would make the claim true.