Assuming only the corners are marked, you certainly can't solve beyond the corners, since you can end up with three 2-cycles of centers.
Short answer to the problem: Yes, there is such a marking system so that we can solve all the corners.
Solution:
Starting with a more specific problem: How far can you get if you mark all corners as 100 where 1 represents white/yellow and 0 represents other colours?
We can reduce it to having all corners oriented with respect to U and D with this marking system. Once you have all the corners oriented on the D layer, you can reduce the corners on the U layer to have at least 2 of the 4 corners oriented.
Case 0: All 4 corners on U are oriented. You're done.
Case 1: 3 of the 4 corners on U are oriented. Put one of the U corners beside the misoriented U corner beneath the misoriented U corner. There's a 50% probability that you can orient all the U corners, and a 50% probability you'll end up with Case 1 again. You can argue that this is never guaranteed to terminate, but c'mon, if you do it enough times you'll surely get it.
Case 2: 2 of the 4 corners on U are oriented. You can reduce it to be the "headlights" pattern as opposed to the "chameleon" pattern. Put down both of the misoriented corners to the D layer with one move each, then you can orient all the U corners.
So it is indeed possible to reduce it to all corners oriented with respect to U and D. These states for the corners can be broken down into three cases.
1. All corners solved,
2. Two diagonal corners swapped on D and two diagonal corners swapped on U.
3. Two U corners on D and two D corners on U.
We can adjust the marking system so that all corners are marked 100 or 110, where white corners are 100 (the white sticker on each corner is marked 1), and yellow corners are marked 110 (the yellow sticker on each corner is marked 0). Then we can use the same approach to orient all the corners and have the white and yellow corners in their correct layer. This would be one of two states, 1. and 2. listed above.
We can adjust the marking system yet again so that we can avoid having a state where all corners are oriented and in their correct layers, but are not permuted correctly. This can be done by marking the white-green-orange corner as 000 and white-orange-blue corners each as 111.
Here's how that works:
Step 1) Complete the white layer so that all of the corners are permuted with respect to each other, and the white-blue-red and white-red-green corners are oriented correctly.
Step 2) Orient as many of the yellow corners as possible.
Case 0: You can orient all of them. You're done -- all the corners are solved.
Case 1: You can only orient three of the four corners.
Case 2: You can only orient two of the four corners.
Step 3) If you had Case 1, one of the two possible D corners are misoriented. Take one of the corners out of the D layer and put it back in twisted clockwise from how it was before. Go back to Step 2.
If you had Case 2, both of the two possible D corners are misoriented. Take both of the corners out of the D layer and put them both back in twisted clockwise from how they were before. Go back to Step 2.
You'll eventually terminate after the white-green-orange and white-orange-blue corners are both oriented correctly, giving a state where all corners are solved.
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