Funnily enough, you will never get to find out what happens when you try to pull it back, because you won't live to see the stick pass through the event horizon. That's not because you will suffer some kind of violent death (although you probably would), it's because you will die of old age before the stick reaches the event horizon. As you push the stick towards the black hole, the subjective time of the end of your stick moves more and more slowly relative to your subjective time.
The Schwartzschild metric can tell us $\frac{d\tau}{dt}$, the rate of time passage at a particular radius $r$ compared to the rate of time passage infinitely far from the black hole:
$$\frac{d\tau}{dt}(r) = \sqrt{1 - \frac{r_s}{r}}$$
where $r_s$ is the event horizon radius. Now, if you are at radius R, and the end of your stick is at radius $R_s$, then the relative rate of time passage of the end of your stick to you is
$$\frac{d\tau}{dt}(R_s) ~/~ \frac{d\tau}{dt}(R) = \frac{d\tau_{stick}}{d\tau_{you}} = \sqrt{\frac{1 - \frac{r_s}{R_s}}{1 - \frac{r_s}{R}}}$$
Notice, that as $R_s$ approaches $r_s$, that ratio approaches zero! By the time your stick is nearly at the event horizon, the end of your stick is experiencing almost no passage of time compared to you.
Now, it seems to me that an interesting question is, if you push on this stick, what do you feel? A force pushing back? What kind of force is it?
When you push on a rigid body like the stick, you are actually sending a wave of pressure through the atoms of the stick that travels at the speed of sound in that material; that's how you effect a force on the front end of your stick without actually touching it. Such a wave would also slow down as it propagates down the length of the stick towards the event horizon, so the front side of the stick would not respond to your force as it normally would. I think that you would experience a kind of "pseudo inertia" - a time-dilation-derived inertia, as though your stick had an enormous mass. But I'll have to think about that some more to be sure.
If the black hole is of sufficient mass, the light would be pulled back towards the black hole
If the light was emitted from outside the event horizon, then the black hole is necessarily not of sufficient mass. Once the photon is emitted, it moves at the local speed of light. It is too late to add mass to the black hole to change its trajectory. The light when it is far from the black hole will be observed to have less energy than an observer at the event would have seen.
the light would be pulled back towards the black hole and would eventually reverse directions.
No, the light never "reverses" directions the way a ball might when thrown upward. Spacetime near the black hole (inside the event horizon) is sufficiently curved that all directions point into the black hole. There is no direction that you can point the light that leads to outside.
Best Answer
One of the problems with describing any situation in general relativity is choosing an appropriate set of coordinates. Far from the black hole we use the Schwarzschild coordinates $t$ and $r$ (we'll ignore the angular coordinates). These are just the radial distance as measured with your rulers and the time measured on your clock so they have a nice simple interpretation. The trouble is that as you approach the event horizon the time gets dilated by a factor:
$$ \frac{d\tau}{dt} = 1-\frac{2M}{r} $$
and at the event horizon, where $r = 2M$, this factor goes to zero. This means time slows to a stop at the event horizon, and it's the source of the common claim that nothing can ever cross the event horizon.
Obviously we're going to struggle to describe what happens to light from a torch inside the event horizon if it takes an infinite time for the torch to even reach the event horizon, let alone cross it. So we need to look for a better coordinate system. We could try using the coordinate system of the astronaut falling in. The trouble with this is that for any freely falling observer spacetime is locally flat, so (ignoring tidal forces) the astronaut thinks they are motionless in flat space. When they turn on the torch the light just speeds off at $c$ as usual.
Outside the event horizon we can use shell observers i.e. observers hovering at a fized distance from the horizon. The trouble is that inside the event horizon it's impossible to hover at fixed $r$, so we can't use shell coordinates either.
So what to do? Well in cases like this we have to choose a set of coordinates that don't correspond directly to anything seen by an observer. This allows us to describe what happens inside the horizon, but at the expense of simplicity. In particular it becomes hard to match the description with our intuitive feel for what happens. Sadly this is a price we have to pay.
The very best coordinates to use are the Kruskal-Szekeres coordinates $u$ and $v$ because they make the causal structure immediately obvious. However these are prohibitively complicated for the non-specialist. The coordinate $u$ is spacelike but isn't simply radial distance, while the coordinate $v$ is timelike but isn't simply time. So I'm not going to use the KS coordinates to answer this question. However, if you're feeling brave have a look at my answers to Would the inside of a black hole be like a giant mirror? and Taking selfies while falling, would you be able to notice a horizon before hitting a singularity? where I use the KS coordinates to answer related questions.
In this case I'm going to use the Gullstrand-Painlevé coordinates. sometimes known as rainfall coordinates or the river model. In these coordinates $r$ is the same radial distance as in the Schwarzschild coordinates, so that's easy to understand. However the time coordinate $t_r$ is the time recorded by a clock carried by a freely falling observer, and because of the time dilation mentioned above this is not the same as the time recorded by the Schwarzschild observer far from the event horizon. Bear this in mind as you consider what follows.
I've already used the GP coordinates to calculate the speed of light heading to or from the black hole in my answer to Why is a black hole black?. The result is:
$$ \frac{dr}{dt_r} = -\sqrt{\frac{2M}{r}} \pm 1 \tag{1} $$
where the $+$ gives the speed of an outgoing ray and the $-$ gives the speed of an ingoing ray. Note that this uses geometric units where $c = 1$. In these units the event horizon is at $r_s = 2GM$. If we use equation (1) to calculate the speed of an outgoing light ray at the event horizon $r = 2M$ we get:
$$ \frac{dr}{dt_r} = -\sqrt{\frac{2M}{2M}} + 1 = 0 $$
and we find that at the event horizon a light ray doesn't shine out, stop and fall back. Instead its velocity is zero so it is fixed motionless and doesn't go anywhere. Inside the event horizon, where $r < 2M $, the velocity of an outgoing ray is negative. So inside the horizon even a light ray directed outwards actually moves inwards not outwards. This is the key result we need to answer the question.
Admittedly we're using an odd time coordinate, but the $r$ coordinate is our good old Schwarzschild coordinate. So while we may may quibble about the exact value of the calculated velocity the sign is unambiguous. That means when our falling astronaut shines his torch outwards the light does not move out, come to a halt and fall back again. The light is moving inwards from the moment it leaves the torch. The reason the astronaut sees the light move away is because the astronaut is falling inwards even faster than the light.
A comment asks if this means the astronaut is moving faster than light, and yes it does. However this shouldn't surprise you as in GR it's only the local velocity of light that is constant at $c$. At distant locations light can move faster or slower than $c$ (though we'll never observe it moving faster as a horizon will get in the way). For example it's well known (or should be!) that sufficiently distant galaxies are moving faster than light.
There is one last loose end to tie up. I claimed above that the astronaut sees the light move away because the astronaut is falling inwards faster than the light is. Can we prove this? It's actually fairly easy to prove if we start from the well known result that the velocity of an observer falling freely from infinity is (in Schwarzschild coordinates):
$$ \frac{dr}{dt} = -\left( 1 - \frac{2M}{r} \right)\sqrt{\frac{2M}{r}} $$
To convert this to the Gullstrand-Painlevé coordinates we note that the rainfall time $t_r$ is just the proper time $\tau$ along the trajectory of the infalling astronaut, and the proper time is related to the coordinate time by the expression I gave above:
$$ \frac{d\tau}{dt} = 1-\frac{2M}{r} $$
The velocity of the astronaut in GP coordinates is then simply:
$$ \frac{dr}{dt_r} = \frac{dr}{dt_r}\frac{d\tau}{dt} = = -\frac{\left( 1 - \frac{2M}{r} \right)\sqrt{\frac{2M}{r}}}{1-\frac{2M}{r}} = -\sqrt{\frac{2M}{r}} $$
Compare this with equation (1) for the velocity of the light, and you'll see that the velocity of light differs from the velocity of the astronaut by $1$. So the light is always moving at a velocity of $c$ relative to the astronaut.