Actually, I don't think it would become a real problem.

Let's assume that it is impossible to have more than 100 leading 0-bits in an SHA-256 hash. When the target becomes 97 leading 0-bits, 1/8 of the eligible hashes are actually impossible. Therefore, the actual difficulty will become 1.14x of the apparent difficulty. 1.33x for 98 bits, 2x for 99 bits, 3.41x for 99.5 bits, 5.33x for 99.7bits, 14.93x for 99.9bits, etc. This will seriously hinder the growth of the apparent difficulty and we may never hit the boundary of 100 bits (unless someone with massive hashing power suddenly push it over the boundary).