This is **not** correct for all primes $p,q;\;$ even  if $p\ne q:\;$  take e.g. $p=2$, $q=3$. Here you have three quadratic residues, two with only two square roots and one with one square root:
$$1^2 \equiv 1 \pmod 6$$
$$ 2^2 \equiv 4 \pmod 6$$
$$3^2 \equiv 3 \pmod 6$$
$$4^2  \equiv 4 \pmod 6$$
$$5^2  \equiv 1 \pmod 6$$