Let me do for balls $\Omega_m$ and I use your Edit. Note that $\bar u_m=\frac{1}{|\Omega_m|} \int_{\Omega_m} u_m \to 0$ by H"older inequality and your assumption. Next I use Poincarè-Wirtinger inequality $$ \|u_m-\bar u_m\|_{p^*} \leq C\|\nabla u_m\|_p $$ with a constant independent of the radius of the ball. This is where I need balls or, more general, dilated domains, since Sobolev inequality is scale invariant for these exponents. If you let $m \to \infty$, you get $\|u\|_{p^*} \leq C$ and then the result follows since $|\Omega_m| \to \infty$. Of course this generalizes to the case where the biggest ball $B_m$ satisfies $|B_m| \geq c|\Omega_m|$ but not to the rectangles you are interested in, so that this is only a partial answer.

Let me do for balls $\Omega_m$ and I use your Edit. Note that $\bar u_m=\frac{1}{|\Omega_m|} \int_{\Omega_m} u_m \to 0$ by H"older inequality and your assumption. Next I use Poincarè-Wirtinger inequality $$ \|u_m-\bar u_m\|_{p^*} \leq C\|\nabla u_m\|_p $$ with a constant independent of the radius of the ball. This is where I need balls or, more general, dilated domains, since Sobolev inequality is scale invariant for these exponents. If you let $m \to \infty$, you get $\|u\|_{p^*} \leq C$ and then the result follows since $|\Omega_m| \to \infty$. Of course this generalizes to the case where the biggest ball $B_m$ satisfies $|B_m| \geq c|\Omega_m|$ but not to the rectangles you are interested in, so that this is only a partial answer.

EDIT Let me complete the proof in the general case. Let $B_m \subset \Omega_m$ be a ball or radius $r_m \to \infty$ and $\bar u_m$ the mean of $u_m$ over $B_m$. If $\bar u_m \to 0$ the argument above gives $u \in L^{p*}$ and the result holds.

If $u_m \to \ell$ (up to a subsequence) then, as above, $u-\ell \in L^{p^*}$ and then $\frac{1}{|\Omega_m|}\int_{\Omega_m}|v|^{p*} \to 0$ with $v=u$ and $v=u-l$. This yields $\ell=0$.