Looks like it is so (though the conclusion is rather that *either $f$, or $g$* is identically $0$ (one of the two is enough).

Let $v$ be the solution of the problem $\Delta v=fg$ in $\Omega$, $v|_{\partial\Omega}=0$. Then, by Green's formula, the integral in question is (up to minus) $\int_{\partial\Omega}u\frac{\partial v}{\partial n}$. However, the boundary values of $u$ can be anything sufficiently nice we want, so this may hold only if $\frac{\partial v}{\partial n}$ is identically zero on the boundary, in which case $v$ can be extended by $0$ outside $\Omega$ and its Laplacian (in the sense of generalized functions, at least) is still $fg\chi_\Omega$.

Now it suffices to show that the Laplacian of a compactly supported function cannot be a product like above unless it is zero. Indeed, its Fourier transform would then be the product $F(z_1)G(z_2,z_3)$ of two entire functions and it should vanish whenever $z_1^2+z_2^2+z_3^2=0$. If there exist $a,b\in\mathbb C$ with $a^2+b^2=-c^2\ne 0$ such that $G(a,b)\ne 0$, then the function $F(cz)G(az,bz)$ of one complex variable vanishes identically and, since the second factor is not zero for $z=1$, we must have $F\equiv 0$, i.e., $f\equiv 0$. Otherwise $G(z_2,z_3)$ is zero on a dense set in $\mathbb C^2$, so it is identically $0$ and so is $g$.