jmc

Since $q(x)$ is quadratic, and we have a horizontal asymptote at $y=0,$ we know that $p(x)$ must be linear.

Since we have a hole at $x=0,$ there must be a factor of $x$ in both $p(x)$ and $q(x).$ Lastly, since there is a vertical asymptote at $x=1,$ the denominator $q(x)$ must have a factor of $x-1.$ Then, $p(x)=ax$ and $q(x)=bx(x-1),$ for some constants $a$ and $b.$ Since $p(3)=3,$ we have $3a=3$ and hence $a=1.$ Since $q(2)=2,$ we have $2b(2-1)=2$ and hence $b=1.$

So $p(x)=x$ and $q(x)=x(x-1)=x^2-x,$ and $p(x)+q(x)=\boxed{x^2}$.