This the classic "hurling a stone into a black hole" problem. It's described in detail in sample problem 3 in chapter 3 of Exploring Black Holes by Edwin F.Taylor and John Archibald Wheeler. Incidentally I strongly recommend this book if you're interested in learning about black holes. It does require some maths, so it's not a book for the general public, but the maths is fairly basic compared to the usual GR textbooks.
The answer to your question is that no-one observes the stone (proton in your example) to move faster than light, no matter how fast you throw it towards the black hole.
I've phrased this carefully because in GR it doesn't make sense to ask questions like "how fast is the stone" moving unless you specify what observer you're talking about. Generally we consider two different types of observer. The Schwarzschild observer sits at infinity (or far enough away to be effectively at infinity) and the shell observer sits at a fixed distance from the event horizon (firing the rockets of his spaceship to stay in place).
These two observers see very different things. For the Schwarzschild observer the stone initially accelerates, but then slows to a stop as it meets the horizon. The Schwarzschild observer will never see the stone cross the event horizon, or not unless they're prepared to wait an infinite time.
The shell observer sees the stone fly past at a velocity less than the speed of light, and the nearer the shell observer gets to the event horizon the faster they see the stone pass. If the shell observer could sit at the event horizon (they can't without an infinitely powerful rocket) they'd see the stone pass at the speed of light.
To calculate the trajectory of a hurled stone you start by calculating the trajectory of a stone falling from rest at infinity. I'm not going to repeat all the details from the Taylor and Wheeler book since they're a bit involved and you can check the book. Instead I'll simply quote the result:
For the Schwarzschild observer:
$$ \frac{dr}{dt} = - \left( 1 - \frac{2M}{r} \right) \left( \frac{2M}{r} \right)^{1/2} $$
For the shell observer:
$$ \frac{dr_{shell}}{dt_{shell}} = - \left( \frac{2M}{r} \right)^{1/2} $$
These equations use geometric units so the speed of light is 1. If you put $r = 2M$ to find the velocities at the event horizon you'll find the Schwarzschild observer gets $v = 0$ and the (hypothetical) shell observer gets $v = 1$ (i.e. $c$).
But this was for a stone that started at rest from infinity. Suppose we give the stone some extra energy by throwing it. This means it corresponds to an object that starts from infinity with a finite velocity $v_\infty$. We'll define $\gamma_\infty$ as the corresponding value of the Lorentz factor. Again I'm only going to give the result, which is:
For the Schwarzschild observer:
$$ \frac{dr}{dt} = - \left( 1 - \frac{2M}{r} \right) \left[ 1 - \frac{1}{\gamma_\infty^2}\left( 1 - \frac{2M}{r} \right) \right]^{1/2} $$
For the shell observer:
$$ \frac{dr_{shell}}{dt_{shell}} = - \left[ 1 - \frac{1}{\gamma_\infty^2}\left( 1 - \frac{2M}{r} \right) \right] ^{1/2} $$
Maybe it's not obvious from a quick glance at the equations that neither $dr/dt$ nor $dr_{shell}/dt_{shell}$ exceeds infinity, but if you increase your stone's initial velocity to near $c$ the value of $\gamma_\infty$ goes to $\infty$ and hence 1/$\gamma^2$ goes to zero. In this limit it's easy to see that the velocity never exceeds $c$.
In his comments Jerry says several times that the velocity exceeds $c$ only after crossing the event horizon. While Jerry knows vaaaaastly more than me about GR I would take him to task for this. It certainly isn't true for the Schwarzschild observer, and you can't even in principle have a shell observer within the event horizon.
Cute idea, +1. Let's think about where the slingshot boost of $2u$ comes from. In the center of mass frame, symmetry guarantees that the test particle exits with a speed equal to the speed with which it entered. If you set this up so that the deflection is nearly 180 degrees, then the problem becomes one-dimensional, so in the c.m. frame, the entry and exit velocities are $v$ and $-v$. Now let's switch to some other frame such as the frame of the sun. This involves adding $u$ to all velocities, so the entry and exit velocities become $v+u$ and $-v+u$. The difference in speed is $2u$.
But this derivation assumed that velocities add linearly when you change frames of reference, which is a nonrelativistic approximation. Relativistically, velocities combine not like $u+v$ but like $(u+v)/(1+uv)$ (in units where $c=1)$. If you put in $v=1$, the result for the combined velocity is always 1.
This is a funny case where we can get the answer to a gravitational problem purely through special relativity. We might worry that the SR-based answer is wrong, because we really need GR for gravity. But we can get the same answer from GR, since GR says that a test particle always follows a geodesic, and a lightlike geodesic always remains lightlike. The reason SR worked is that an observer could watch a patch of flat space far away from the black hole, observe a wave-packet of light passing through that patch on the way to the black hole, and then observe it again on the way back out. Since the patch is flat, SR works.
The patch argument also justifies using the SR equation for the Doppler shift to find the effect on the energy and momentum of the scattered wave. This effect happens without a change in velocity. The black hole recoils, and the total energy-momentum is conserved.
Best Answer
To answer your question you need to be clear what coordinates you're using. If you use coordinates that are co-moving with the rock falling into the black hole then the rock will see the event horizon pass at the speed of light.
External observers, using Schwarzchild coordinates, will see the rock slow down as it approaches the horizon, and if you wait an infinite time you'll see it stop.
External observers obviously can't comment on the speed of the rock after it has passed the event horizon because it takes longer than an infinite time to get there. If you use the rock co-moving coordinates then you can ask what speed you hit the singularity and ... actually I'm not sure what the answer is. I'll have to go away and think about it.
Incidentally http://jila.colorado.edu/~ajsh/insidebh/schw.html is a fun site describing what happens when you fall into a black hole.