By interpolating "legal configuration" if it were to exist, I think you can obtain a nowhere vanishing, continuous vector field on $S^2$. Suppose that for some $n$ there is a legal configuration on the $n\times n \times n$ cube. For each 1x1 face, put a vertex at the center. Then connect vertexes whose 1x1 face's touch. (basically something like a dual graph, but I don't know what the real terminology is).
Doing this, you get $(n-1)^2$ squares on each face of the Rubik's cube, $n-1$ squares on each edge of the Rubik's cube, and one triangle for each vertex of the cube. Now, put the arrow from the 1x1 face at the associated vertex. For each square or triangle we can now linearly interpolate to get a vector field over the whole thing. These will patch back together to form a continuous vector field on $S^2$ because on the lines they are glued along, the value on each piece is linear interpolation between the same two vectors. Thus, basically it remains to check that given a "legal configuration" you cannot interpolate to a zero vector.
The squares are not too bad, because the most that two of the vectors being interpolated can be off by is $90^\circ$, so you can't get a zero vector.
The triangles are a little trickier because things get twisted around, but if you try to write down the possible cases, you can see that there is basically only one type of "legal" corner configuration, and it doesn't interpolate to a zero vector.
This seems to show that you don't even need to assume anything special about corner squares, you can allow them to point in the 8th illegal position, and there are still no "legal configurations."
There is another stable solution of total length $3\pi + h + \epsilon$ for any $\epsilon > 0$. We take a circle of constant latitude $\delta < 0$ (sufficiently small) and then connect this circle to the north pole via two diametrically opposed strings. This is then clearly stable. Furthermore this is equivalent to your solution (plus Scott Carnahan's epsilon-modification).
However, there is in fact a stable solution of total string-length $2\pi+h+\epsilon$. We simply take the almost-equatorial circle in the last solution and drag it to the south pole, so that we have a small circle there along with two diametrically opposed support strings connecting it to the north pole. This solution is (barely) stable, although physically it is not exactly easy to implement (a very small but non-negligible disturbance will suffice to remove the ball).
The reason for the two answers: this was my thought process in action.
EDIT: Please disregard the above; both of these solutions are unstable.
In fact Scott's example is part of a general class of solutions of total length $3\pi+h+\epsilon.$ Take any two diametrically opposite points, and draw a small spherical triangle containing one of the points. Connect all six possible edges between the four points by geodesics, and finally rotate the arrangement so that one of the strings passes through the north pole.
Here is a short proof that any stable configuration must have at least four points where three or more strings meet. If there are three or less points, there is a hemisphere $H$ which contains all of the points. Take the complement $H^C$ of this hemisphere; we can remove any strings in $H^C$ because they cannot be geodesics. As $H^C$ doesn't contain the north pole, the sphere can fall out.
Note that if any strings are not geodesics between junctions, we can ignore the strings.
Here is a short proof (that is not quite rigorous) that could provide a lower bound. Suppose there is not a loop (that is, a set of points connected by geodesics) that is completely contained in the southern hemisphere. Then we may drag any points in the southern hemisphere into the northern hemisphere. (This statement is the part I can't make completely rigorous.) So the sphere must be unstable. Now if we have a loop in the southern hemisphere, there must be at least two strings meeting at the north pole, as otherwise we could simply slide all of the string off one side of the sphere.
So we must have a loop in the southern hemisphere and (not necessarily direct) connections from at least two of these points to the north pole. I can't figure out how to work a good lower bound from here.
Best Answer
Consider trajectories that start on the slope. We measure success by how long it takes to reach any point a given vertical distance below the starting point (which is not exactly the success measure in the question but may be a satisfactory replacement).
Let $s$ denote distance along the trajectory. Let $t(s)$ be the time at which point $s$ is reached, and $y(s)$ be the vertical displacement from the starting point (measured downwards). Also let $u$ be the initial speed. Since the collisions are elastic, conservation of energy implies that the speed of the ball when it has gone distance $s$ along its trajectory is $\sqrt{u^2+2g\, y(s)}$. From this it follows that $$t(s) = \int_{0}^s \frac{dq}{\sqrt{u^2 + 2g\, y(q)}}.$$ Now suppose you could choose any trajectory lying on or above the smooth slope. For trajectories of given length $s$, you can minimise $t(s)$ by maximising $y(q)$ as much as possible for $0\le q\le s$, which seems a bit vague value until you realise that moving down the slope makes $y(q)$ take its maximum possible value $q/\sqrt 2$ for all $q$. Since this straight line also minimises the length of the trajectory to any given vertical distance below the starting point, we can infer a theorem:
The staircase can be drawn so that the inner corners lie on the slope. If you have to start at such an inner corner, the above theorem shows that you can't choose any initial starting speed and direction so that the ball will reach a given vertical distance below the starting point faster than it would if you used the same initial speed down the smooth slope.
If you can start at places other than the inner corners, a bit more work needs to be done (starting with an exact definition of the problem). But it is still clear that the slope is better eventually.
A further, much harder, question is how much worse the staircase is. As Ryan commented, repeated glancing collisions seem best but I'm not sure if such trajectories are physical. Might it be that no matter how hard you try eventually the ball will encounter hard collisions that make it bounce upwards? This would be disastrous since it can bounce up to almost the same height it started from.
ADDED: In the above I treated the ball as a point mass. If it is a disk of finite radius (but still frictionless so the angular inertia is not an issue), replace what I called "the slope" by the lowest 45-degree line that the centre of the disk may touch and "the trajectory" by the path taken by the centre of the disk. Draw the staircase so that the centre of the ball is on that line when the ball is sitting on a step as far in as possible. The analysis appears to be the same.