Will the black hole size increase?
First I will describe a few facts about entropy and how it applies to black holes, and then I will apply these facts to your question.
Entropy is observer-dependent. Entropy is (roughly speaking) the logarithm of the number of microstates compatible with the system's macroscopic state. What the macroscopic state is depends on what you can measure. This is true even for mundane things like mixtures of gases - in this case the entropy simply depends on what instruments you have available for making measurements. (See Jaynes ,1996.)
In the case of an object falling in to a black hole, the entropy depends on whether you're falling in with it or watching from the outside. If you're watching from the outside, time dilation means that any radiation emitted by the object will rapidly be redshifted to unobservably low frequencies. This means that any information about the object besides its mass, charge and angular momentum is, for all practical purposes, lost. The object is still technically there, since nothing can pass the event horizon as seen from the perspective of an outside observer, but since you can't see it any more, the black hole's macrostate is said to consist only of its mass, charge and angular momentum. The number of microstates associated with a black hole's macrostate rapidly increases with its mass, which means that the entropy increases by quite a lot when something falls into it, when seen from outside.
However, for someone falling in with the object it's a different story. They can still see the object, so they don't lose any information about it, and its entropy doesn't do anything special as it passes the event horizon. The entropy increases for the outside observer but stays the same for the falling observer. There is no paradox, because entropy is observer-dependent.
We now have enough information to answer your question. From your point of view, falling into the black hole with your partitioned box of gases, nothing much happens when you pass the event horizon. Then you mix the gases, and watch their entropy increase just as it normally would. Nothing particularly special seems to happen as a result. Some time later, you and the box of gas hit the singularity and suddenly cease to exist.
From my point of view as an observer staying safely outside the black hole, it looks a bit different. I watch you falling towards the event horizon with your box of gas, but as you approach it you seem to move in slow motion. Your light gets red-shifted until it gets harder and harder to see you, and at some point I just give up and say you've become part of the black hole, which now has a greater mass and hence a higher entropy and a larger radius. (The radius and the entropy are always proportional to one another.)
In principle, if I could see you, I would see you frozen in time just before you passed the horizon; the gases would still be unmixed. Nothing you do after passing the horizon can ever have an effect on what I observe. But if you'd mixed the gases before passing the horizon it wouldn't have made any difference. The black hole's macrostate depends only on its mass, charge and angular momentum, and those are all the same whether the gases are mixed or separate. In general, an increase in entropy is a loss of information, but by the time you've fallen into the black hole I've already lost all the information about the gases' microstate, so mixing them together can't cause me to lose any more. So from the outside viewer's perspective, the radius of the black hole increases when you fall into it, and then that's it. The entropy is already at its maximum at this point, and there's nothing you can do that can increase it further.
This is an interesting question, and I think ultimatly relates to the black hole information paradox. The first law of black hole thermodynamics states that $$TdS = dE - dW$$ where the variables have their usual meaning. From this one can derive what is called the Bekenstein-Hawking entropy $$S_{BH} = \frac {c^3 A} {4 G \hbar}$$ where $A$ is the surface area of the black hole. As correctly pointed out by JamalS, in the Schwarzschild geometry this takes the form: $$S = 4 \pi G M^2$$ What this is in effect saying is that only the $\textit{mass}$ of an object that has the unfortunate fate of bonding with the black hole will contribute to its entropy. So, while in a classical system you are correct in saying opening the box increases the entropy of the system, the information contained in the partition is scrambled as it crosses the horizon. As such, only the combined mass content of you and the box contribute to the entropy regardless of whatever atomic structure the two of you may have contained.
For more complicated black hole models the surface area may contain terms involving the angular momentum, $J$, and the net charge, $Q$ with electric potential $\Phi$, and the term $dW$ in the first equation has the form $$dW = \Omega dJ + \Phi d Q$$ However, the classical black hole no hair theorems state that the black hole is uniquely characterised by $M$, $Q$ and $J$. This is what is called 'no-hair', since no other information contained in the infalling particles is retained. When one introduces some strange Yang-Mills fields (exotic matter) and things into the scenario the black hole may exhibit some hair (see e.g [http://cds.cern.ch/record/312731/files/9610019.pdf]). So, one could argue that there may exist exotic states of matter that when they fall into the black hole they maintain some of their internal information. In light of this, the Bekenstein-Hawking formula would need a correction term and you may indeed be able to increase the entropy (and consequentially surface area) by having some clever partition set up like you describe provided the internal atoms consist of exotic matter.
Edit: I may have messed up some of the multiplicative constants in the above, but the picture is the same.