Does there exist bivariate polynomials $p$ and $q$ such that $p(x,y)^2 = q(x, y)^2 ( x^2 + y^2)$? Does there exist bivariate polynomials $p$ and $q$ such that $p(x,y)^2 = q(x, y)^2 ( x^2 + y^2)$ for all real $x$ and $y$?

The answer is no.
If $p(x,y)^2 = q(x, y)^2 ( x^2 + y^2)$ for all real $x$ and $y$, then $p(X,Y)^2 = q(X,Y)^2 (X^2 + Y^2)$ in $\mathbb R[X,Y]$. 
Since $X^2+Y^2$ is irreducible (hence prime) in $\mathbb R[X,Y]$ we get $p(X,Y)=(X^2+Y^2)p_1(X,Y)$, and therefore $(X^2+Y^2)p_1(X,Y)^2=q(X,Y)^2$. Using the same argument we have $q(X,Y)=(X^2+Y^2)q_1(X,Y)$, and thus we get $p_1(X,Y)^2 = q_1(X,Y)^2 (X^2 + Y^2)$. Obviously, $\deg_X p_1<\deg_Xp$ and $\deg_X q_1<\deg_Xq$. Continuing the procedure we obtain a relation of one of the above forms, say $p(X,Y)^2 = q(X,Y)^2 (X^2 + Y^2)$ with $\deg_Xp=1$ and $\deg_Xq=0$. Can you continue from here?
Remark. A slight generalization: in a UFD is not possible to have a relation of the form $a^2=b^2p$ with $a,b$ non-zero and $p$ prime.
