Hollow Cylinder vs Solid Cylinder Thanks to angular momentum, we know that hollow cylinders are slower than solid cylinders when rolled down an inclined plane - Is this difference in speed (or time taken to get to the bottom of the incline) noticeable with a full and empty can of deodorant? 
Is there any factor that would determine whether this difference would be noticeable (such as length of incline, angle of incline, etc.)? 
How could variables be changed so that this difference is noticeable?
 A: The short answer is "yes".  
As you say, "we know that hollow cylinders are slower than solid cylinders when rolled down an inclined plane".  That is, a solid cylinder will roll down the ramp faster than a hollow steel cylinder of the same diameter (assuming it is rolling smoothly rather than tumbling end-over-end), because moment of inertia depends on the distribution of mass, with mass further from the axis of rotation contributing more to moment of inertia than mass closer to the axis.
But that's only part of what's in play here.  Here we have the can, which is rolling, and the fluid, which is basically just falling.  The assumption I'm making is that the fluid has a viscosity close to that of water, since the deodorant is mostly water, and by simply shaking a can of it we can tell it's not terribly viscous - it shakes like water in a can.
I just did an experiment, rolling two identical transparent bottles, one empty, the other about half full, down about a 15-degree ramp.  The empty bottle behaved as you would expect an empty cylinder to, rolling slower than the full bottle.  What was also clear is that for a low-viscosity fluid like water, there wasn't enough friction for the water to "spin up" within the bottle - it all stayed on the bottom half of the bottle throughout the journey, so the water added nothing to the moment of inertia about the bottle's axis.  However, being much more massive than the bottle, the water contributed much more to the acceleration than did the bottle itself.  Also, friction between the water and the bottle wasn't enough to make much difference.  Had the liquid been much thicker, the result would be much different, but in this case, the difference in speed between the two bottles was obvious.
For the latter part of your question, the inclined plane is what forces the can to rotate - if you just dropped them, moments of inertia wouldn't factor in at all, and the full and empty cans would drop at very close to the same rate.  The shallower the ramp is, the more the can has to rotate for a given drop in height, and so the more clear the difference is between a can whose motion is dominated by angular momentum (empty can) versus another whose motion is dominated by linear momentum (full can).
A: This really depends on a couple different things, If the deodorant is a gas or if it is solid, what you mean by noticeable and what type of friction there is. 
For the first part if you are comparing a solid deodorant than the solid deodorant can will always be sqrt(3)/2 faster than the empty can (a simple conservation of energy problem). 
If it is a gas than it gets more complicated. The effect will be much less noticeable as it is will be harder to create rotation among the gas to the point if you consider the gas to be an ideal gas then there will be almost no difference. 
To make this effect more noticeable is to increase the height that the cans are released from. This means either increasing the length of the ramp or increasing the angle. Mass and radius of the cylinders should not matter.
If the amount of friction is only enough to cause the cylinders to roll without slipping then all I said above is true but if you have less friction then the cylinders will just slid down the ramp and have the same speed. If you have more friction then you will have a much more complicated problem. 
A: The difference is called mass moment of inertia. For the general case of cylinder with an inner and outer wall the formula is
$$ I_{\rm cyl} = \frac{m}{2} \left( r_O^2 + r_I^2 \right) $$
The two extreme cases are:


*

*Solid cylinder - $r_O = R$, $r_I = 0$ $$I_{\rm cyl} = \frac{m}{2} R^2$$

*Hollow cylinder - $r_O = R$, $r_I=R$ $$I_{\rm cyl} = \frac{m}{2} 2 R^2 = m R^2$$


Now take a hollow cylinder and add a gas (no liquid) inside that rotates with the cylinder (steady state solution) then the gas will add to the mass moment of inertia and it will slow down the cylinder (accelerate less) under the same conditions.
The total MMOI is:
$$I_{\rm total} = I_{\rm cyl} + I_{\rm gas} = m_{\rm cyl} R^2 + \frac{m_{\rm gas}}{2} R^2 $$
So in theory yes, but in reality since $m_{\rm gas} \ll m_{\rm cyl}$ you won't notice a difference.
