Useless mathematician answer: Take $Q$ and $R$ to be the same group and take any $P$ that intersects trivially with them.
But, more seriously, here is an example from $S_5$. Let $Q$ and $R$ be the $2$-Sylow subgroups $Q=\langle (1,2),(1,3)(2,4)\rangle$ and $R=\langle (4,5),(1,4)(2,5)\rangle$. These intersect nontrivially: $Q \cap R = \langle (1,2) \rangle$. Now, let $P=\langle (3,4),(1,4)(3,5)\rangle$ and $P \cap Q = 1$ as required.
Edit: Just for fun I found the smallest group which has a counterexample of the type you mention. It's $S_3 \times S_3$. If we represent the group by $S_3 \times S_3=\langle (1,2,3),(1,2),(4,5,6),(4,5)\rangle$, we've got $P=\langle (1,2),(4,6) \rangle$, $Q=\langle (2,3),(5,6) \rangle$, and $R = \langle(2,3),(4,5)\rangle$. This is probably a better example anyway.