Chopping off SHA-256 entropy?
Assuming the hash function is good, and SHA-256 is widely believed to be so, the probability of no collisions in $k$ samples into a range of size $n$ (e.g., obtained by selecting a subset of characters) is upper bounded by $exp(-k^2/2n).$

For you, say $k=2^{21}=2,097,152$ (couple million) and $n=2^{64}$ means that your probability of no collision is roughly

$$exp(-2^{42}/2^{65})=exp(-2^{-21})\approx 1-1.2\times 10^{-7}.$$

If you had 8 times more inputs (16 million plus) this would become

$$ exp(-2^{48}/2^{65})=exp(-2^{-17})\approx 1-7.6\times 10^{-6}$$

so probability of collision would be more than one in a million.