Your intuition is that of Newtonian mechanics, I can say this because you said "Black holes have such a strong gravity that escape velocity from it is more than speed of light which basically means nothing could escape it". In fact, gravity is much more complicated than that. In the frame of general relativity, it is not surprising to have a Black-Hole whose surface gravity is less than Earth's.
As your question seems to focus only on the horizon's gravity, I will not consider the internal structure of a static black hole solution in General Relativity. But I do think that it is noteworthy to consider it too in order to have a good understanding of a black hole. Furthermore, I will say "infinite amount of" because the maths blow up at the coordinate singularity (the event horizon) in the coordinates I chose. But it is to be interpreted as "when you approach the horizon from the exterior, things tend to either $0$ or $\infty$".
Firstly, what is an event horizon in General Relativity, in simple words? To keep things simple, I will only consider Schwarzschild's black holes, which do not rotate. An event horizon is a surface in space such that the first component of the metric tensor is precisely $0$. To understand this statement, let us focus on the Schwarzschild line element, which we can interpret as a deviation of the geometry of space-time from the Minkowski picture:
\begin{equation}
ds^2=\underbrace{\left(1-\frac{r_s}{r} \right)}_{\text{radial time dilation}}\times\overbrace{dt^2}^{\begin{array} \\ \text{infinitesimal}\\ \text{time parameter}\end{array}}-\underbrace{\left( 1-\frac{r_s}{r} \right)^{-1}}_{\text{radial length contraction}}\times\overbrace{dr^2}^{\begin{array} \\ \text{infinitesimal} \\ \text{radial parameter} \end{array}}-\underbrace{r^2d\Omega_2^2}_{\text{angular part}}
\end{equation}
Concretely, $ds^2$ is the "square" of an infinitesimal length in space-time and is called the "space-time interval" (among other appellations). Note that when $r=r_s$, the "radial time dilation" becomes 0, this means that the event horizon is located at a radius $r_s$ from the origin center. Of course, the "radial length contraction" factor blows up at this radius, but this is a feature of Schwazrchild black holes. In the Kerr geometry, which describes a rotating black hole, this is no longer the case.
But what does it mean to have a time dilation factor of $0$? It means that, for the faraway observer, times will appear to stop at the horizon. So if you are far away from the horizon and drop an apple towards it, the apple will gradually stop moving, and ultimately be completely stopped right at the horizon. Of course from the point of view of the apple, time does not stop. But if the apple has a camera stuck to it that films you, it will see you in accelerated motion. Ultimately, when the apple is at the horizon, it would film the death of the universe (of course, if the camera has an infinite lifetime, and if the black hole does not evaporate).
Now, I have focused on the falling of the apple only, but not on its appearance. As you may know, you see electromagnetic radiations from ~400nm to ~800nm. But, as the apple falls towards the horizon, the light it scatters in your direction will be redshifted. So light still does travel at $c$, but the more it is emitted near the horizon (and directed outwards), the more its wavelength is dilated. Ultimately if the light, which has a non-zero wavelength, is emitted toward the faraway observer from the horizon, then from the point of view of this very observer${}^{\ast}$, light has an infinite wavelength, and so a frequency of $0$, and so does not exist anymore. Thus, black holes are classically... well, black.
I still must clarify some things about your question. The phenomenon you describe well, the fact that objects having a speed below the escape velocity fall back to the Earth, happens for light too, but when you are inside of the black hole!
Also, what I described is purely in terms of the faraway coordinates $(t,r,\theta,\varphi)$. In other coordinate systems, you have other conclusions. Among those is, for example, the fact that you can fall in a finite time into the black hole. But in this case, the time parameter is not the cosmological time $t$ anymore. See this Wikipedia link for the other representations of the line element I wrote above.
${}^{\ast}$Finally, I said that you will no longer see the emitted light, but this is if you stay still for an infinite amount of time! Indeed, due to the "radial length contraction" term, which blows up at the horizon, light has to travel an infinite distance to reach you, and so it takes an infinite amount of time.
