This conjecture seems to be true:
If the product of two different primes is a sum of two squares, the so are the primes.
I've tested it for $pq<1000$, but would like to see a proof.
Edit: If $p\neq q$ are primes and $pq=a^2+b^2$, then there are integers $c,d$ such that $p=c^2+d^2$.
This is true by fermat's sum of two squares theorem.
the theorem states that an integer $n$ is a sum of two squares if and only if every prime $p\equiv \bmod 3$ that divides has an even exponent.
Using this:
If $pq$ is a sum of two squares we cannot have $p\equiv 3$ or $q\equiv 3$.
So both $p$ and $q$ are sums of two squares.