No, it is not possible to get exactly equal probabilities for a deterministic mapping from all 256-bit numbers to the range 0-99.
However, you can ask whether it matters, since a bias on the order of $2^{-256}$ is undetectable. A mapping that took the 256-bit number modulo 100 and refused the inputs less than $2^{256} \bmod 100$ would be unbiased and would never fail in practice.
You can also get the next best thing – probabilities that are not exactly equal, but for which no one knows which of the numbers 0-99 is biased which way. For example, you can define the mapping using another SHA-256 iteration like so:
- Take the SHA-256 hash of the initial hash concatenated with the number 1 (e.g. in ASCII): $H(h||1)$. If this is at least $2^{256} \bmod 100$, return it modulo 100.
- Otherwise increment the counter and calculate $H(h||2)$, doing the same check.
- Continue as long as necessary (i.e. in practice you never need to go further than the first step).
Since we don't know which $h$ (if any) produce $H(h||1) < 36$, we don't know the resulting $H(h||2)$.