Here is my answer. I should preface it by warning that this is a subject that can provoke intense discussion, and I'm sure there are physicists would would disagree. You should be aware that I'm an expert on thermodynamics but not on general relativity.
But basically, as far as I understand it, the process of converting matter into black-hole-stuff is an irreversible one, in the usual macroscopic sense. Throwing your boxes of salt into identical black holes is somewhat analogous to what would happen if you emptied them into two identical vats of water. You would end up with two identical vats of salty water, with the same mass, temperature, and salt concentration, and the same entropy.
The no hair theorem for black holes is an asymptotic one. It says that if you throw some stuff into a black hole and wait long enough, the black hole will become an arbitrarily good approximation to an "ideal" black hole (which is to say, a black hole solution of Einstein's equations), which can can be completely described by its mass, charge and spin. It also says (I believe) that this convergence happens rather rapidly. But converging towards something is not the same as ever actually reaching it. In reality, nothing can cross the event horizon as seen from an outside perspective (see my answer to this question), it just gets very hard to detect because its light is red-shifted to extremely long wavelengths.
So in my view the apparent loss of information comes from assuming that the black hole actually becomes an ideal one rather than just closely approximating it. It's very similar to the question of how the entropy of an isolated vat of salt+water can increase as the salt dissolves, even though on the microscopic level, the laws of physics seem to preserve information. The resolution is that when you switch to a macroscopic description (in terms of temperature, pressure etc.), you throw away some information about the microscopic state. After the salt has dissolved, the information about its previous state (crystal or powder) is still there, but it's hidden in fine correlations between the molecules' motions. When you choose to describe the final state as an equilibrium ensemble you're basically admitting that those fine correlations can never practically be measured, and therefore choosing to ignore them. Similarly, when you choose to approximate a real black hole as an ideal one, you're basically choosing to ignore any information about what kind of salt was thrown into it in the past, on the basis that there's no longer any practical way to recover it. In both cases, the fundamental reason for the increase in entropy is the same.
Note that I'm not saying the box's entropy increases as it passes the event horizon. I'm actually saying that the box never crosses the event horizon, as seen from an outside point of view. That would take an infinite amount of time. However, the outside observer would very rapidly find the box very hard to see due to the red-shifting. At some point you, as the observer, might decide as an approximation that the box might as well have crossed the event horizon, since you basically can't detect it anymore. When you do this, your approximation has a higher entropy than the "real" black hole, and that's where the entropy increase comes from.
That might seem like a weird concept. But in fact all increases in entropy are due to approximations of one kind or another. In principle you could always reverse the velocities of every particle making up a system and watch it "run backwards in time" to its initial state (unscrambling an egg or whatever). So the information about the initial conditions is always still there. We just treat things as irreversible (i.e. information-destroying or entropy-producing) because it's a very useful approximation that helps us make predictions about macroscopic systems.
Of course, the observer falling in with the box of salt would not want to make the same approximation as the outside observer. It would be a bad approximation from the infalling observer's point of view, because she can still see the box perfectly clearly. (If it's a big enough black hole it won't even get torn apart.) But that's ok - although we often treat it as an observer-independent physical quantity, entropy is actually observer-dependent, even for everyday things like gases. See this rather wonderful paper by Edwin Jaynes. (Jaynes, E. T., 1992, `The Gibbs Paradox, ' in Maximum-Entropy and Bayesian Methods, G. Erickson, P. Neudorfer, and C. R. Smith (eds.), Kluwer, Dordrecht).
These are two different situations.
When an ordinary object falls into a black hole, The black hole curves spacetime. The object is assumed to not curve spacetime. A typical solution is given by the Schwarzschild metric. In that solution, as seen by a distant observer, the object takes an infinite time to reach the event horizon.
When a small black hole is dropped in, both objects curve spacetime. The small black hole is not going to get arbitrarily close to the big event horizon without disturbing it. The metric between the black holes is not well described by the Schwarzschild metric. They will merge.
Best Answer
Yes it is, because null areas (and only null areas) are invariant to the time-slicing, although it isn't completely obvious to someone who isn't day-to-day familiar with Minkowki geometry. This is proved here in the first titled section of this answer: Second Law of Black Hole Thermodynamics
Perhaps I should say how this proves the intended result: no matter what spacelike surface you slice a stationary black hole with, so long as matter didn't fall in (so that the geodesics on the black hole surface didn't spread apart), the area you cross with the slice is invariant. A boosted observer has a tilted simulteneity plane, so this observer crosses the horizon with a different natural coordinate (although which natural coordinate to use is not specified by GR--- you can use isotropic Schwarzschild coordinates, then boost these by a naive Lorentz transformation (this keeps the asymptotic metric flat), and then continue a little bit into the interior by any method, this requires bending the t-coordinate away from the symmetry direction so that it can cross the horizon). The area of each little triangle on the crossing surface can be slid up and down to match each little triangle on another crossing surface, so the area is independent of the frame.