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: $$1^2 \equiv 1 \pmod 6$$ $$ 2^2 \equiv 4 \pmod 6$$ $$3^2 \equiv 3 \pmod 6$$ $$4^2 \equiv 4 \pmod 6$$ $$5^2 \equiv 1 \pmod 6$$

This is not correct for all primes $p,q;\;$ even if $p\ne q:\;$ take e.g. $p=2$, $q=5$. Here you have two quadratic residues in $(\mathbb{Z}/n\mathbb{Z})^\times$ namely $1$ and $9\equiv -1,\;$ but both have only two square roots: $$1^2 \equiv 9^2 \equiv 1 \pmod {10}, \quad \text{and}\quad 3^2 \equiv 7^2 \equiv 9 \pmod {10}$$