Take the quotient of $R$ by $P$, then localized at the image of $Q$. You then have a noetherian local ring of dimension $\ge 2$. Any prime ideal of height $1$ of this local ring will induce a prime ideal of $R$ strictly included between $P$ and $Q$, and two such prime ideals will induce two distinct prime ideals.
Now in a noetherian ring of dimension $\ge 2$, there are always infinitely many prime ideals of height $1$, see my answer to this question.
As user26857 answered the question in a comment, and prefers not to post it as an answer, I'll try to write the answer myself. I think I've understood user26857's argument, but I may be wrong. So, in the lines below, everything that's true is due to user26857, and everything that's false is due to me.
The answer is Yes.
More precisely:
If $A$ is a noetherian integral domain, if $\mathfrak p_1,\dots,\mathfrak p_k$ are distinct nonzero prime ideals of $A$, and if $m$ and $n$ are distinct elements of $\mathbb N^k$, then we have $$\mathfrak p_1^{m_1}\cdots\mathfrak p_k^{m_k}\ne\mathfrak p_1^{n_1}\cdots\mathfrak p_k^{n_k}.$$
Proof. In the setting of the question, suppose by contradiction that we have
$$
\mathfrak p_1^{m_1}\cdots\mathfrak p_k^{m_k}=\mathfrak p_1^{n_1}\cdots\mathfrak p_k^{n_k}
$$
with $m\ne n$.
Enumerate the $\mathfrak p_i$ in such a way that each $\mathfrak p_i$ is a minimal element of the set $\{\mathfrak p_i,\dots,\mathfrak p_k\}$, and write $\mathfrak p_{ij}$ for the localization of $\mathfrak p_i$ at $\mathfrak p_j$.
For all $i$ we get
$$
(\mathfrak p_{1i})^{m_1}\cdots(\mathfrak p_{ii})^{m_i}=(\mathfrak p_{1i})^{n_1}\cdots(\mathfrak p_{ii})^{n_i}.\quad(1)
$$
Note the following consequence of the determinant trick, or Nakayama's Lemma:
$(2)$ If $\mathfrak a$ and $\mathfrak b$ are ideals of $A$, then the equality $\mathfrak a\mathfrak b=\mathfrak b$ holds only if $\mathfrak a=(1)$ or $\mathfrak b=(0)$.
Let's prove $m_i=n_i$ by induction on $i$:
Case $i=1$: We have $(\mathfrak p_{11})^{m_1}=(\mathfrak p_{11})^{n_1}$ by $(1)$. If we had $m_1\ne n_1$ we could assume $m_1<n_1$, and would get
$$
(\mathfrak p_{11})^{n_1-m_1}(\mathfrak p_{11})^{m_1}=(\mathfrak p_{11})^{m_1},
$$
contradicting $(2)$.
From $i-1$ to $i$: We have
$$
(\mathfrak p_{1i})^{m_1}\cdots(\mathfrak p_{i-1,i})^{m_{i-1}}(\mathfrak p_{ii})^{m_i}=(\mathfrak p_{1i})^{m_1}\cdots(\mathfrak p_{i-1,i})^{m_{i-1}}(\mathfrak p_{ii})^{n_i}.\quad(3)
$$
If we had $m_i\ne n_i$ we could assume $m_i<n_i$ and we could write $(3)$ as
$$
(\mathfrak p_{ii})^{n_i-m_i}\mathfrak b=\mathfrak b
$$
with $(\mathfrak p_{1i})^{n_i-m_i}\ne(1)$ and $\mathfrak b\ne(0)$, contradicting $(2)$. (Here $\mathfrak b$ is the left side of $(3)$, and we assume $2\le i\le k$.) $\square$
Note that the argument still works if $A$ is not noetherian, but the $\mathfrak p_i$ are finitely generated.
Best Answer
Consider the collection of nontrivial ideals of $R$ which do not contain a product of nonzero prime ideals. If this collection is nonempty, then it contains a maximal element $I$. I claim that $I$ must be prime (which is absurd).
Suppose otherwise, so that there exist $x,y \notin I$ such that $xy\in I$. The ideals $I+xR$, $I+yR$ are strictly larger than $I$, and they are not trivial: if, say, $I+xR=R$ then $yI+yxR = yR$ but $yI+yxR$ is contained in $I$, contradicting that $y \notin I$. Therefore by the choice of $I$ each of them must contain a product of prime ideals, but then so does $(I + yR)(I+yR) \subseteq I$, which contradicts the choice of $I$.