I think that this is 16/21, or about 76.19%, but I'm not completely certain that I haven't missed some esoteric parity complication. Needs more thought...
Okay, I've given the parity issue more thought, and have determined that I was right to doubt my first impression. Parity excludes some of the scenarios I was thinking were possible. If we assume that the observer is clever enough to calculate the corner parity of a scrambled cube in his head, then the result increases to 6/7, or about 85.714%.
Given a scrambled 2x2 that is stickered the standard way, one can see all three sides of the UFL and UFR pieces, so we know their permutation and orientation. Similarly, we see two sides each of the UBR, DFR, UBL, and DFL pieces, and since we know that the color scheme is standard, we can determine the permutation and orientation of them as well. This leaves only the DBL and DBR pieces, of which we only see one sticker each. Since we know by process of elimination what pieces occupy those two positions, we need only figure out their permutation (2 possibilities) and orientation (3 possibilities taking parity into account).
There are three main scenarios for these two semi-hidden pieces. The first is that they are corners that share no colors, i.e. opposite corners. Each piece only has one such counterpart out of 7 available, so this scenario has a 1/7th chance of occurring. In this case, knowing the parity state of the rest of the cube, seeing one sticker is enough to determine both the orientation and permutation of both pieces. We can see two stickers, so in this scenario there is zero chance of confusion.
The second scenario is that the two pieces share exactly one color, i.e. they are corners that belong on the same face but are otherwise opposite. Each piece has 3 such counterparts out of 7 available, so this scenario has a 3/7ths chance of occurring. In this case, any sticker that isn't the one color they share will instantly give the game away, so only the orientation where the common color is exposed on both pieces will work. In 2/3rds of these cases, parity will make this arrangement impossible, and in the remaining third, there is a 1/3rd chance that the common color will be the one exposed. (1/3)*(1/3)*(3/7)=1/21.
The third and final scenario is that the two pieces share exactly two colors, i.e. they belong adjacent to each other. Let us call the two colors they have in common color A and color B. As before, there are three parity cases. In one case, an orientation exists that reveals stickers A and B, but not one that reveals stickers B and A. In another parity case, an orientation exists that reveals stickers B and A, but not one that reveals stickers A and B. (This bit is what tripped me up the first time 'round.) It seems like you would be able to confuse them, but if you have perfect understanding of parity, it should be possible to determine which corner is which just by seeing A and B or B and A along with all the other stickers that are in view.
This leaves a third and final parity state where either A and A can be in view, or B and B can be in view, or the distinct colors can be in view, with equal probability. A and A will confuse, as will B and B, and so there is a 2/3rds chance of confusion, but only a 1/3rd chance of this parity state. Multiply in the probability of having exactly two colors in common, and you get (2/3)*(1/3)*(3/7)=2/21.
Add them all up and you get 0+1/21+2/21=1/7. A 1 in 7 chance of the cube state being confusable with another, or a 6/7 chance of it being uniquely identifiable merely by viewing the U, F, L, and R sides.
Mind you, this is just for question A. Question B, which I believe allows for non-standard color schemes, is one I am unprepared to answer.
