Why does Hawking radiation cause black holes to die? If a particle is being expelled from a black hole and an antiparticle is being driven into it, shouldn't the opposite occur as well and in the same frequencies?
I mean, black holes should emit antiparticle radiation as well and gather energy from it and the black hole energy shouldn't change, right?
What am I missing?
 A: EDIT:The full machinery of quantum field theory in curved space-time is required to derive Hawking radiation properly. In QFT there are various equivalent ways to think about scattering processes. The best and easiest way, due to Feynman diagrams, is to think about virtual particles. These particles are not real, and in this sense they do not exist. They are counterparts of real particles, but they can have any mass they like. They "exist" only as intermediaries between real particles. The popular science explanation of Hawking radiation mostly focuses on this virtual particle picture. What does this have to do with vacuum energy? Well, if you have a field and it has the lowest possible energy (vacuum energy) how many particles there are is not a "natural" question, so to speak. It is something that can change for short periods of time, due to the famous Heisenberg Uncertainty Principle. So these virtual particles are really just fluctuations around the vacuum energy, but note that the negative energy of one of the particles is absolutely necessary in Hawking radiation, because at the end of this process you need to have a situation which is real and is not just a fluctuation. The energy of the vacuum can not be lowered (it is the lowest possible energy the field can have) so the energy has to come from the black hole. The vacuum is not what gives the energy! If there were no vacuum there would be no fields, and no Universe, as we know it, not just no Hawking radiation. :)
People usually talk about event horizon and not the ergosphere because when they explain Hawking radiation they focus on the simpler case of a black hole which does not rotate. See this answer about Hawking radiation of rotating, charged black holes.
OLD ANSWER: Black holes do emit all sorts of different particles and antiparticles as a part of Hawking radiation. The loss of mass has nothing to do with whether matter or antimatter falls into a black hole. It has to do with the sign of the energy of the (virtual) particle. 
Quantum fluctuations create particle-antiparticle pairs near the event horizon, but these are virtual. That means that they can't exist very long and can only be observed indirectly. One of these (anti)particles manages to escape and become a real particle with positive energy. Conservation of energy demands that the (anti)particle which did not escape and has fallen into the black hole has negative energy. The opposite does not happen because if it were to happen the negative energy virtual particle would become a real particle and this is not possible. 
A: The following answer is not "rigourous", but it may gives a simple explanation.
Suppose you have a quantum fluctuation just near the horizon, but outside.
This quantum fluctuation create 2  particles, one with a negative energy -E, the other with a positive energy  +E.
If the 2 particles stay outside the black hole, they have to anihilate themselves in a time time $t \le \frac{\hbar}{E}$
Now, one of the 2 particles may fall inside the black hole, and there are 2 possibilities ; the escaping particle may have a positive or a negative energy.
The key point is that there is an asymmetry between these 2 cases.
For a particle to be real, its energy has to be positive, but relatively to the time coordinate. With an evolution variable $\tau$, this can be write $\frac{dt}{d\tau}>0$ 
When the horizon is being crossed (by the infalling particle), we may consider, that there is a change in the nature of the time and radial space coordinate. The time-like coordinate becomes a space-like coordinate, and the radial space coordinate becomes a time-like coordinate.
More precisely, if, outside the black hole, the coordinate are(in units $c=1$) : $z=r+it$, then the "coordinates" inside the black hole are $z \rightarrow z'\sim -iz$
So, $$z'=r'+it'\sim-i(r+it)=(t-ir)$$
So, $t'\sim-r$ and $x'\sim r$
For an escaping particle, the energy must be positive relatively to $t$, so $ E=\frac{dt}{d\tau} >0$, but for the infalling particle the "energy" must be positive relatively to $t'$, that is = $\frac{dt'}{d\tau}>0$, which is "equivalent" to  $-\frac{dr}{d\tau}>0$.
But the last expression means only that the particle is an infalling particle, which was our hypothesis. We could say also, for the infalling particle, that the "outside" energy $-E$ becomes a "inside" momentum.
A: It is called energy conservation.
When a real particle departs from the field of the black hole, energy is taken away to infinity, and the hole is depleted by that delta(E), no matter how soft the particle leaving is. Eventually it is possible that all the mass  of the black hole will be depleted to the point of not being able to have  the  strength of the gravitational field that makes it a black hole .

Now if you are worried that the universe is in danger of losing its black hole population don’t be. Hawking radiation only has a noticeable effect on black holes of around 10^12 kg 1. or smaller. This is because a black hole must be starved to death. That is the matter-energy it puts out must be greater than the matter-energy it takes in. Since black holes mass is inversely proportional to its temperature and the amount of matter-energy that it radiates is dependent on its temperature, then as the mass of the black hole goes up the amount of matter-energy that it emits goes down. So for larger back holes such as one about the mass of our sun their temperature is very low about 10^-7 K and so they have very low radiation. And since they receive light from stars, cosmic back ground radiation, and even matter, in the form of dust, planets, and stars they continue to grow. It will only be at some future date when all stars have gone out, and matter is out of reach of large black holes and when the temperature of the universe is less than that of the black hole itself that the larger black holes will begin to evaporate.

Edit after comments:
The above replies to the title.
In the content you ask:

If a particle is being expelled from the Black Hole and an antiparticle is being driven into it, shouldn't the opposite occur as well and in the same frequencies?

Yes.

I mean Black holes should emit antiparticle radiation as well and gather energy from it and the black hole energy shouldn't change, right?

No.
Whether it is a particle or an antiparticle, the energy it carries, mass and kinetic part, is always positive for a real particle. Once a particle/antiparticle leaves the black hole it is real and carries away energy depleting the store of energy of the black hole.

What am I missing?

That antiparticles are anti only in their quantum numbers versus their corresponding particles, not in the energy/mass. The electron and the positron both have the same positive mass.
A: I was the one who posted the question... But I'm having HUGE problems with the website as I cannot comment anything and I cannot log in into eMagus account for some reason (everytime I login into eMagus it automatically logs into eJunior). So I cannot reply to any answer unless I post it as an answer as well... Sorry.
I have done some research and the problem I'm having with your answers are this: There is no such thing as a negative energy particle; it is impossible to exist. Both the antiparticle and particle are positive in energy.
They 'draw' energy from the vacuum... And normally they would collide and release the energy back in the vacuum. I suppose that Hawking radiation theory has nothing to do with negative energy particles... Its origins must be from the vacuum energy drained comiong from the black hole so no matter if the particle or antiparticle are erradiated, the 'energy drained' comes from the microparticles from the vacuum of the event horizon.
But what I don't understand is why the energy vacuum radiation drained comes from the event horizon and NOT from the ergosphere of the black hole?
In my mind it is the vacuum from the ergosphere that should provide energy to the radiation and not the vacuum from the black hole itself. If just barely two virtual particles from the liminar of the ergosphere are created and one fall into the event horizon, the energy still comes from the ergosphere, right?
And even if it were true that is draining energy from the ergosphere, what I already have trouble understanding, shouldn't the ergosphere emmit Hawking radiation also and on a much bigger scale than anything that could possibly come from the event horizon?
I'm trying to merge the two accounts to comment properly, thank you!
