I would imagine that, the energy that pulls the quark pair into the black hole is intrinsic energy (ie it comes from the black hole), and the quarks created through 'spaghettification' would only be proportional to that energy pulling on them. Therefore, seeing as energy is equal to mass and vice versa this new mass would not be created, rather transformed from the energy of the black hole. The black hole never gains more mass/energy than the original quark pair because it already contained the mass/energy in its system that would be then created into more quark pairs.
I hope that makes sense
I suspect that what has confused you is the difference between remaining a fixed distance from the black hole and falling freely into it. Let me attempt an analogy to illustrate what I mean.
Suppose you are carrying a large and heavy backpack. You can feel the gravitational force of the backpack weighing you down. However this only happens because you're staying a fixed distance from the centre of the Earth i.e. you're standing stationary on the Earth's surface. If you and the backpack were to leap from a cliff then (ignoring air resistance) you would feel no gravity as you plummeted downwards and the backpack wouldn't weigh anything.
If we now switch our attention to the black hole, if you attempt to stay a fixed distance from the black hole (presumably by firing the rocket motors on your spaceship) you'd feel the weight of the backpack, and the weight would get bigger and bigger as you approach the event horizon. In fact the weight is given by:
$$ F=\frac{GMm}{r^2}\frac{1}{\sqrt{1-\frac{r_s}{r}}} \tag{1} $$
where $m$ is the mass of the backpack, $M$ is the mass of the black hole, $r_s$ is the event horizn radius and $r$ is your distance from the centre of the black hole. As you approach the event horizon, i.e. as $r \rightarrow r_s$, equation (1) tells us that the force goes to infinity. That's why once you reach the event horizon it is impossible to resist falling inwards.
But you only feel this force because you're trying to resist the gravity of the black hole. If you just fling yourself off your spaceship towards the black hole then you will feel no weight at all. You would fall through the event horizon without noticing anything special. In fact you would see an apparent event horizon retreating before you and you would never actually cross anything that looks like a horizon to you.
But there is another phenomenon that can cause you problems, and this is related to the phenomenon of spaghettification that you mention. At any moment some parts of you will be nearer the centre of the black hole than others. For example if you're falling feet first your feet will be nearer the centre than your head. That means your feet will be accelerating slightly faster than your head, and the end result is that you get slightly stretched. This is called a tidal force, and it happens with all sources of gravity, not just black holes. Even on the Earth the gravitational force on your feet it slightly higher than on your head, though the difference is so small that you'd never notice it.
The thing about a black hole is that because its gravity is so strong the tidal forces can get very strong indeed. In fact they can get so strog that they'd pull you out into a long thin strip like a piece of spaghetti - hence the term spaghettification.
But the tidal forces only become infinite right at the centre of the black hole. They are not infinite at the event horizon, and in fact for large enough black holes the tidal forces at the event horizon can be negligably small. The equation for the variation of gravitational acceleration with distance is:
$$ \frac{\Delta a}{\Delta r} = \frac{c^6}{(2GM)^2} \tag{2} $$
If we take a black hole with the mass of the Sun and use equation (2) to calculate the tidal force we get $\Delta a/\Delta r \approx 10^{9}g$/m. So if you're two metres tall the difference between the acceleration of your head and feet would be $2 \times 10^9g$, where $g$ is the gravitational acceleration at the Earth's surface. This would spaghettify you very effectively. However at the event horizon of a supermassive black hole with the mass of a million Suns the difference between your head and feet would be only 0.001$g$ and you'd struggle to feel it.
Best Answer
It's not an assumption, it's a calculation plus a theorem, the Penrose singularity theorem.
The calculation is the Tolman-Oppenheimer-Volkoff limit on the mass of a neutron star, which is about 1.5 to 3 solar masses. There is quite a big range of uncertainty because of uncertainties about the nuclear physics involved under these extreme conditions, but it's not really in doubt that there is such a limit and that it's in this neighborhood. It's conceivable that there are stable objects that are more compact than a neutron star but are not black holes. There are various speculative ideas -- black stars, gravastars, quark stars, boson stars, Q-balls, and electroweak stars. However, all of these forms of matter would also have some limiting mass before they would collapse as well, and observational evidence is that stars with masses of about 3-20 solar masses really do collapse to the point where they can't be any stable form of matter.
The Penrose singularity theorem says that once an object collapses past a certain point, a singularity has to form. Technically, it says that if you have something called a trapped lightlike surface, there has to be a singularity somewhere in the spacetime. This theorem is important because mass limits like the Tolman-Oppenheimer-Volkoff limit assume static equilibrium. In a dynamical system like a globular cluster, the generic situation in Newtonian gravity is that things don't collapse in the center. They tend to swing past, the same way a comet swings past the sun, and in fact there is an angular momentum barrier that makes collapse to a point impossible. The Penrose singularity theorem tells us that general relativity behaves qualitatively differently from Newtonian gravity for strong gravitational fields, and collapse to a singularity is in some sense a generic outcome. The singularity theorem also tells us that we can't just keep on discovering more and more dense forms of stable matter; beyond a certain density, a trapped lightlike surface forms, and then it's guaranteed to form a singularity.
This question amounts to asking why we can't have a black-hole event horizon without a singularity. This is ruled out by the black hole no-hair theorems, assuming that the resulting system settles down at some point (technically the assumption is that the spacetime is stationary). Basically, the no-hair theorems say that if an object has a certain type of event horizon, and if it's settled down, it has to be a black hole, and can differ from other black holes in only three ways: its mass, angular momentum, and electric charge. These well-classified types all have singularities.
Of course these theorems are proved within general relativity. In a theory of quantum gravity, probably something else happens when the collapse reaches the Planck scale.
Observationally, we see objects such as Sagittarius A* that don't emit their own light, have big masses, and are far too compact to be any stable form of matter with that mass. This strongly supports the validity of the above calculations and theorems. Even stronger support will come if we can directly image Sagittarius A* with enough magnification to resolve its event horizon. This may happen within 10 years or so.