Electric field outside a capacitor

Outside two infinite parallel plates with opposite charge the electric field is zero, and that can be proved with Gauss's law using any possible Gaussian surface imaginable. However, it might be extremely hard to show if you don't choose the Gaussian surface in a smart way.

The usual way you'd show that the electric field outside an infinite parallel-plate capacitor is zero, is by using the fact (derived using Gauss's law) that the electric field above an infinite plate, lying in the $xy$-plane for example, is given by $$ \vec{E}_1=\frac{\sigma}{2\epsilon_0}\hat{k} $$ where $\sigma$ is the surface charge density of the plate. If you now put another plate with opposite charge, i.e. opposite $\sigma$, some distance below or above the first one, then that contributes its own electric field, $$ \vec{E}_2=-\frac{\sigma}{2\epsilon_0}\hat{k} $$ in the region above it. Since the electric field obeys the principle of superposition, the net electric field above both plates is zero. The same happens below both plates, while between the plates the electric field is constant and nonzero.

Your way of doing it is a little more tricky, but again gives the same answer. For example, if you choose the Gaussian surface to have an hourglass shape with different radii for the two sides, then indeed the net charge enclosed is not zero. However, when you calculate the total electric flux through that surface, you have to be careful to realize that there is nonzero electric field between the two plates, and therefore there is a nonzero flux through the part of the Gaussian surface that lies between the plates. That flux, of course, has to be accounted for. Assuming that you know the electric field inside the capacitor, $\vec{E}_\text{inside}$, you can do the integral $\oint\vec{E}_\text{inside}\cdot d\vec{A}$ for such a Gaussian surface (it's not that hard actually), and you find that the flux through the part of the surface that lies between the plates is exactly equal to $q_{\text{enclosed}}/\epsilon_0$. Thus, the net flux through the part of the Gaussian surface that lies outside the plates has to be zero, proving, after a little thought, that the electric field outside the capacitor is zero.

The final answer for $\vec{E}$ never depends on the Gaussian surface used, but the way to get to it always does. That's why the Gaussian surface has to be chosen in a smart way, i.e. in a way that makes the calculation of $\oint\vec{E}\cdot d\vec{A}$ easy.

You can pick such a box. However, this box alone does not allow you to determine the electric field. Gauss' law states that the total flux is equal to the charge enclosed times $4\pi$. It doesn't tell you where that flux leaves the box. There could be a lot here, none there, a negative amount at a third place, etc.

In some situations, symmetry allows us to use Gauss' law to find the electric field. A Gaussian box enclosing both plates of a parallel-plate capacitor is symmetric with respect to a reflection through a plane through the middle of the capacitor. It is also symmetric with respect to rotations about an axis perpendicular to the plates (ignoring edge effects). The rotational symmetry lets us say that the electric field first points along the axis perpendicular to the plates. The reflection symmetry tells us that the electric field must be the same through both sides of the box parallel to the plates. That electric field must be zero because the box has no net charge in it.

A box around a single plate does not have the same symmetries. It still has the rotational symmetry, so the electric field must be perpendicular to the plates. However, we don't have the reflection symmetry any more, so the electric field can be different strengths on different sides of the box.

We know there must be net flux through the box because the box encloses net charge. Using the fact that the electric field is zero outside the capacitor, we can deduce the he flux through a box that encloses only one plate is all through the side of the box that's inside the capacitor. Hence, the electric field must be $4\pi\rho$ inside the capacitor.