I do not think Gauss theorem is necessary to solve the equation. We can have the answer just with integration by part: \begin{align} \int_{\Omega}{(\partial_{x_1} u\partial_{x_1} \phi+\partial_{x_2} u \partial_{x_2} \phi)d^2x} & = \int{\bigg(\int{(\partial_{x_1} u)d\phi\big|_{x_2}}\bigg) dx_2}+\int{\bigg(\int{(\partial_{x_2} u)d\phi\big|_{x_1}}\bigg) dx_1} \\ & = -\int_{\Omega}{\bigg({\partial^2 u \over \partial x_1^2}\bigg)\phi d^2x}-\int_{\Omega}{\bigg({\partial^2 u \over \partial x_2^2}\bigg)\phi d^2x} \\ & = -\int_{\Omega}{(\nabla^2u)\phi d^2x} \end{align} If we plug this back in the equation derived from the Hamiltonian's principle, we have $$\int_{t_1}^{t_2}{\int_{\Omega}{(-\rho\ddot{u}+\sigma\nabla^2u)\phi d^2x}dt}=0$$ This gives us the wave equation $-\rho\ddot{u}+\sigma\nabla^2u=0$.
Edit: The folowing is the updated version of how Gauss theorem can be applied to the problem. I did not think of the function $\nabla \cdot (\phi \nabla u)$ in my original consideration.
At last, let us take a look how Gauss theorem can be applied to the problem, and you will find why I think it is unnecessary. Since $$\nabla \cdot \nabla (u\phi) = \phi\nabla^2u+u\nabla^2\phi +2\partial_{x_1} u\partial_{x_1} \phi+2\partial_{x_2} u \partial_{x_2} \phi$$$$\nabla \cdot (\phi \nabla u) = \nabla\phi \cdot \nabla u +\phi\nabla^2u = \partial_{x_1} u\partial_{x_1} \phi+\partial_{x_2} u \partial_{x_2} \phi+\phi\nabla^2u$$ the integral $\int_{\Omega}{(\partial_{x_1} u\partial_{x_1} \phi+\partial_{x_2} u \partial_{x_2} \phi)d^2x}$ is \begin{align} \int_{\Omega}{(\partial_{x_1} u\partial_{x_1} \phi+\partial_{x_2} u \partial_{x_2} \phi)d^2x} & = {1 \over 2}\int_{\Omega}{(\nabla \cdot \nabla (u\phi)-\phi\nabla^2u-u\nabla^2\phi)d^2x} \\ & = {1 \over 2}\bigg(\int_{\partial \Omega}{\nabla (u\phi) \cdot dl}-\int_{\Omega}{(\nabla^2u)\phi d^2x}-\int_{\Omega}{u\nabla^2\phi d^2x}\bigg) \end{align}\begin{align} \int_{\Omega}{(\partial_{x_1} u\partial_{x_1} \phi+\partial_{x_2} u \partial_{x_2} \phi)d^2x} & = \int_{\Omega}{(\nabla \cdot (\phi \nabla u)-\phi\nabla^2u)d^2x} \\ & = \int_{\partial \Omega}{(\phi \nabla u) \cdot dl}-\int_{\Omega}{(\nabla^2u)\phi d^2x} \end{align} The first term, $\int_{\partial \Omega}{\nabla (u\phi) \cdot dl}$$\int_{\partial \Omega}{\phi \nabla u \cdot dl}$ is $0$ since $\phi=0$ on the boundary, and the third term, with integration by part, is \begin{align} \int_{\Omega}{u\nabla^2\phi d^2x} & = -\int_{\Omega}{(\partial_{x_1} u\partial_{x_1} \phi+\partial_{x_2} u \partial_{x_2} \phi)d^2x} \\ & = \int_{\Omega}{(\nabla^2u)\phi d^2x} \end{align} Therefore, we have $\int_{\Omega}{(\partial_{x_1} u\partial_{x_1} \phi+\partial_{x_2} u \partial_{x_2} \phi)d^2x}=-\int_{\Omega}{(\nabla^2u)\phi d^2x}$. However, with Gauss theorem, we still cannot avoid the use of integration by part, so it seems unnecessary. $$\int_{\Omega}{(\partial_{x_1} u\partial_{x_1} \phi+\partial_{x_2} u \partial_{x_2} \phi)d^2x}=-\int_{\Omega}{(\nabla^2u)\phi d^2x}$$