Is collision resistant hash enough for one-block counter mode?
It depends what you mean by collision resistance.
If by collision resistance you mean that the output of $H(.)$ cannot be distinguished from a stream with full collision entropy (i.e. a $H_2=\ell$) then by Jensen's inequality we cannot distinguish from a stream with Shannon entropy at least this large and so the conditions for a one-time pad are met.
If by collision resistance you mean that hash function collision cannot be found without less work than $n$ hash function evaluations, then no. Let $n=2^{256}$, let $\ell=1024$ and define $H(x)$ to be SHA3(x)||SHA3(x). Collisions in $H(x)$ are in one-to-one correspondence with SHA3 collisions and so cannot be found with fewer that $2^{256}$ hash evaluations (with the proviso that SHA3 remains strong). Thus there is a 256-bits of collision resistance, but the encryption function is now horribly weak as the every other 512-bit block of key stream is a repeat of the previous.