I won't answer the questions one after another but will try to share my form of understanding from which all the answers should be clear.
First of all, imagine what happens when you charge the capacitor. You take two parallel plates and connect each of them by a wire to one pole of a battery. Current from the source flows for a short time, not really going through the capacitor, but in a way that looks similar from the outside: electrons flow to plate A and plate B gives out other electrons which flow back to the other pole of the battery, keeping the chemical reaction in it going. (It wouldn't work well if more electrons left one pole than came back into the other.)
Since the capacitor plates are split by an insulator, the charge of the electrons accumulates on plate A. Opposite charge of the same absolute amount arises on the other plate, since we pulled electrons out of it. Both the plates contribute to an electric field which is strongest and almost homogeneous in the are between the two plates. This creates a potential difference, which happens to be in the same direction as the battery's EMF, thus there's less reason for more current to charge the capacitor.
The process stops when the voltage across the capacitor matches the battery's EMF. (This would strictly speaking take an infinite amount of time but nothing's perfect anyway.) Now you have two charged plates of opposite signs. The plates are not the only charged thing, the rest of the circuit, which is also conductive, is still connected to them. However, there's a slight difference, as the electrons on the surface of plate A facing the plate B are attracted by the Coulomb force of the positive charge of plate B. The electrons covering the rest of the negatively charged surface are repelled from each other and from the aforementioned area, too. We can effectively neglect them and say that only the side of the plate A facing plate B contains negative charge and vice versa.
Now we come to the numbers. Some negative charge -$Q$ is distributed over an area $S$ on plate A and a positive charge $+Q$ is distributed over an area $S$ on plate B. This immediately gives that the area charge density is $-Q/S$ or $+Q/S$, respectively. The next step to get us closer to the relation between charge $Q$ and the corresponding voltage $U$ is Gauss' law. It states that the area integral of $E$ over any surface equals $1/\epsilon$ of the total charge of the volume enclosed the surface, i.e. inside. This really says that any charge outside the surface has zero net effect on the integral (answering your 3.)! How can this be the case? It's because the flow lines created by outside charges enter the surface but leave it later. This contributes to the surface integral by both signs. And the particular form of Coulomb's law is mathematically so very clever that these exactly cancel out. So only charges located inside work, because they don't share this behaviour.
So, let us take some reasonable surface. If we encompassed the whole capacitor in it, the net charge inside is zero, which would render Gauss somewhat useless. If we only cut parts of the plates or took parts of the surface not parallel to them, this would just mean extra work. So we naturally take only one of the places, whole, and a bit of free space around it (so that it's really clear what is inside and what is not). Let us say we take the positive plate. The charge inside is $+Q$. The total surface of the box is somewhat bigger than two times the area $S$. However, we remember that almost all of the electric field is acting between the places, perpendicular to them, and is homogeneous in space. So there will only be some electric field crossing our surface on an area of size $S$ which really lies "between" the plates, and the $E$ will be more or less constant and with at a right angle over this area. Thus, Gauss says that
$$E.S = \frac{Q}{\epsilon}.$$
Finally, the voltage, dealing with a homogeneous field, is just $E.d$, giving our desired relation defining capacity,
$$U = E.d = \frac{d}{S}\cdot\frac{Q}{\epsilon} \quad \Rightarrow \quad Q = \frac{\epsilon S}{d} U =: C.U.$$
You might wonder why I assumed above that the field behaves the way it does. This was an important point in using the law. Well, we can try to reach the answer without this assumption, using only what we know about sheets of charge, thus deciding not to use Gauss' law. For this purpose, let the plate A be a sheet of negative charge and plate B a sheet of positive charge. We know that either of them generates the $E$ field of strength $\frac\rho{2\epsilon}$ on its either side. This seems to be half of what I said above, and moreover also out of the capacitor. But there's actually no contradiction, just a different way of reaching the same answer.
For an easier description, let the plates are vertical, plate A is on the left and plate B on the right. Plate A is negative, thus creates a field of strength $\frac\rho{2\epsilon}$ pointing to the right on the left of the capacitor, to the left inside the capacitor and to the left to the right of it, too. Plate B is positive, so its field points away from it in all the three regions. It turns out that the two $\frac\rho{2\epsilon}$ are in different directions and cancel out when outside of the capacitor, but they help each other and sum up to $\frac\rho\epsilon$ inside. This is the same result Gauss' law gave us.
Remember that in this way, there is no selection of volume and its enclosing surface and all that stuff, we must consider everything. Gauss' law is another way, equally correct. It gives the same answer usually more quickly at the cost of the need of more assumptions or better said, intuition. Feel free to choose what suits you better, but be prepared to need both, eventually.
Final remark on the charges being exactly opposite: well, apart from the fact that there are fluctuations and such which would make the balance inexact anyway, there actually does not need to be any. I used a capacitor involved in an electrically neutral circuit. However, you may very well take it out, charge one plate using, say, a van der Waals generator, and not let the charge carriers charge leave the other plate. In this case, the charges on the two plates are not of the same size. However, Coulomb forces act which repel the charge of the same sign as on plate A away from plate B, concentrating it on other parts of the conductor's surface and inducing an opposite charge on the facing side. The charge on A also redistributes. As a result, the capacitor would behave as if it was neutral with charges $\frac Q2$ and $-\frac Q2$ on its plates (in terms of both potential difference and electric field), with an ambient electric field of a discharged capacitor with equal charges on both sides superimposed to it. You could measure this part of the field outside of the capacitor. However, when a capacitor is used in circuits, which is the most common use one may need to deal with, no net charge of it as a whole changes its properties in any way, so we always just safely and silently ignore this possibility.
The mistakes you are doing are:
You didn't considered the flux coming from them in between them. You have to take all the flux in all directions coming from them. You should take the gaussian across the surface of the plane otherwise you will get wrong result.
$|\vec E_+|=|\vec E_-|=\frac{\sigma}{2\epsilon_0}$ and not $\frac{-\sigma}{2\epsilon_0}$ for $|\vec E_-|.\space$ $\sigma$ is the magnitude of the charge density.
You are incorrectly adding the fields which gave you $0$ inside. The magnitudes have to be added when directions are same and subtracted when directions are opposite.
This is what we get from Gauss's law: $$\vec{E}=\frac{\sigma}{2\epsilon_0}\hat r$$
where, $$|\vec{E}|=\frac{\sigma}{2\epsilon_0}$$where $\sigma$ is the magnitude of surface charge density
So, outside, if direction of $\vec{E_+}$ is $\hat r$ then, direction of $\vec{E_-}$ is $-\hat r$ $$\vec{E_+}=\frac{\sigma}{2\epsilon_0}\hat r$$$$\vec{E_-}=\frac{\sigma}{2\epsilon_0}(-\hat r)$$$$\vec{E_+}+\vec{E_-}=\frac{\sigma}{2\epsilon_0}\hat r+\frac{-\sigma}{2\epsilon_0}\hat r$$ $$=0$$
Inside, both $\vec E_+$ and $\vec E_-$ has same direction $\hat r$ $$\vec{E_+}=\frac{\sigma}{2\epsilon_0}\hat r$$$$\vec{E_-}=\frac{\sigma}{2\epsilon_0}\hat r$$$$\vec{E_+}+\vec{E_-}=\frac{\sigma}{2\epsilon_0}\hat r+\frac{\sigma}{2\epsilon_0}\hat r=\frac{\sigma}{\epsilon_0}\hat r$$
Talking in magnitudes, inside, the magnitudes have to be added, $$|\vec E_+|+|\vec E_-|=\frac{\sigma}{2\epsilon_0}+\frac{\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}$$
outside, they have to be subtracted, $$|\vec E_+|-|\vec E_-|=\frac{\sigma}{2\epsilon_0}-\frac{\sigma}{2\epsilon_0}=0$$
Best Answer
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.