Solution:
Yes. Let $t$ be an arbitrary positive integer and consider the following perfect square:
$$(t!)^{2} = (t!) \cdot (t! - 1)! \cdot t!.$$
So if we consider $(p,q,r) = (t!, t! - 1, t)$ then $p!q!r! = (t!)^{2}$ which is a perfect square. Since the choice of $t$ is arbitrary, there must be infinitely many such triples.

Solution:
Yes. Consider the substitution $(p,q,r) = (6t^{2}, 6t^{2} - 1, 3)$
$$p! \cdot q! \cdot r! = (6t^{2})! \cdot (6t^{2} - 1)! \cdot 3! = ((6t^{2} - 1)! \times 6t)^{2}$$
which is a perfect square. Since the choice of $t$ is arbitrary, there must be infinitely many such triples.

Solution:
This solution shows a stronger result. For any positive integer $k$, there exist infinitely many $q$ such that $(q + 1)! \cdot q! \cdot k$ is a perfect square. This can be seen by setting $q + 1 = kx^{2}$ for any $x$. This of course implies the required result.