jmc

The polynomial $P(x)\cdot R(x)$ has degree 6, so $Q(x)$ must have degree 2. Therefore $Q$ is uniquely determined by the ordered triple $(Q(1),Q(2),Q(3))$. When $x=1$, 2, or 3, we have $$0=P(x)\cdot R(x)=P\left(Q(x)\right).$$It follows that $(Q(1),Q(2),Q(3))$ is one of the 27 ordered triples $(i,j,k)$, where $i$, $j$, and $k$ can be chosen from the set $\{1,2,3\}$.

However, the choices $(1,1,1)$, $(2,2,2)$, $(3,3,3)$, $(1,2,3)$, and $(3,2,1)$ lead to polynomials $Q(x)$ defined by $Q(x)=1$, $2,$ $3,$ $x,$ and $4-x$, respectively, all of which have degree less than 2. The other $\boxed{22}$ choices for $(Q(1),Q(2),Q(3))$ yield non-collinear points, so in each case $Q(x)$ is a quadratic polynomial.