It is true that, from an outside perspective, nothing can ever pass the event horizon. I will attempt to describe the situation as best I can, to the best of my knowledge.
First, let's imagine a classical black hole. By "classical" I mean a black-hole solution to Einstein's equations, which we imagine not to emit Hawking radiation (for now). Such an object would persist for ever. Let's imagine throwing a clock into it. We will stand a long way from the black hole and watch the clock fall in.
What we notice as the clock approaches the event horizon is that it slows down compared to our clock. In fact its hands will asymptotically approach a certain time, which we might as well call 12:00. The light from the clock will also slow down, becoming red-shifted quite rapidly towards the radio end of the spectrum. Because of this red shift, and because we can only ever see photons emitted by the clock before it struck twelve, it will rapidly become very hard to detect. Eventually it will get to the point where we'd have to wait billions of years in between photons. Nevertheless, as you say, it is always possible in principle to detect the clock, because it never passes the event horizon.
I had the opportunity to chat to a cosmologist about this subject a few months ago, and what he said was that this red-shifting towards undetectability happens very quickly. (I believe the "no hair theorem" provides the justification for this.) He also said that the black-hole-with-an-essentially-undetectable-object-just-outside-its-event-horizon is a very good approximation to a black hole of a slightly larger mass.
(At this point I want to note in passing that any "real" black hole will emit Hawking radiation until it eventually evaporates away to nothing. Since our clock will still not have passed the event horizon by the time this happens, it must eventually escape - although presumably the Hawking radiation interacts with it on the way out. Presumably, from the clock's perspective all those billions of years of radiation will appear in the split-second before 12:00, so it won't come out looking much like a clock any more. To my mind the resolution to the black hole information paradox lies along this line of reasoning and not in any specifics of string theory. But of course that's just my opinion.)
Now, this idea seems a bit weird (to me and I think to you as well) because if nothing ever passes the event horizon, how can there ever be a black hole in the first place? My friendly cosmologist's answer boiled down to this: the black hole itself is only ever an approximation. When a bunch of matter collapses in on itself it very rapidly converges towards something that looks like a black-hole solution to Einstein's equations, to the point where to all intents and purposes you can treat it as if the matter is inside the event horizon rather than outside it. But this is only ever an approximation because from our perspective none of the infalling matter can ever pass the event horizon.
It's tidal forces that pull the object apart. The key point is that there isn't a local inertial frame covering the whole object. This is by definition - we're talking about an extended object in the question!
To get an intuition for what's going on it's more helpful to split the object into several smaller pieces, each of which have an approximate local inertial frame. For simplicity we'll just consider two objects, and suppose they are joined by some kind of rope. This is just a simple model for nearby atoms of the material held together by interatomic forces.
Say object $A$ is on a geodesic which escapes the black hole and object $B$ is on which which falls in. The first thing to remember is that $A$ will never actually see $B$ reach the horizon due to time dilation in his frame of reference! This is obvious if you draw the setup in Kruskal coordinates.
But at some point your rope will break. This is because the proper distance between the observers must grow as object $B$ falls towards the singularity and $A$ escapes away from the black hole. In each of the frames of $A$ and $B$ this will manifest as a tugging force on the rope which is eventually too large for the rope to bear.
It's worth noticing that this tidal force could be arbitrarily small when $B$ crosses the horizon if $A$ and $B$ started close and the black hole is large. It will only become apparent later on.
In conclusion I think that the event horizon was a bit of red herring here. As usual it's tidal forces which rip things apart. This doesn't have to be instantaneous just because part of your object has passed through the horizon!
For more details, including calculations see this article by Greg Egan. He actually considers the analogous (and less terrifying) scenario where a Rindler horizon is created by an accelerating observer. But the mathematical ideas can be carried over to this setup.
Best Answer
With the proper definition of the meaning of gravitational acceleration, the questioner is correct and the other answers that claim that the gravitational force at the event horizon is infinite are wrong.
Per Wikipedia:
So with this definition, the OP is correct that the suitably defined surface gravity at the event horizon decreases as the mass of the black hole increases.
Now this surface gravity does not mean that a rocket engine that can produce that acceleration will enable you to hover at that distance from the black hole. It does take an infinitely powerful rocket engine to hover arbitrarily close to the horizon and, of course, no rocket engine could let you hover inside the event horizon.
However, if both observers, A and B are freely falling in from infinity, nothing at all unusual will happen as first B and then A (one meter later) crosses the event horizon. Neither will lose sight of the other at any time. What actually happens is that the photons bouncing off of B as he crosses the horizon will be frozen at the horizon waiting for A to run into them at the "speed of light". B who is inside can toss the ball to A who is falling in but is currently outside the event horizon and A will catch the ball after he crosses the event horizon. This is true since to first order A and B, when freely falling are in a common inertial reference frame and they can do whatever they could do when far from the black hole.
The problem comes if they try to hover with one person inside and one outside the horizon. That is not possible – the person inside cannot hover at all and the person outside would need a very powerful continuously firing rocket engine to try to hover. But then all the effects of time dilation etc. will be occurring for both of them and all the problems noted by the other answers will be true.
Read these questions and answers for more insight:
How does an object falling into a plain Schwarschild black hole appear from near the black hole?
How does the star that has collapsed to form a Schwarschild black hole appear to an observer falling into the black hole?