Is collision resistant hash enough for one-block counter mode? Let $H:\{0,1\}^*\rightarrow \{0,1\}^\ell$ be a collision resistant hash function (CRH), and let $\Pi = (\mathsf{enc,dec})$ be an encryption scheme with the following properties:

$\Pi.\mathsf{enc}: \mathcal{K} \times \{0,1\}^\lambda \rightarrow \{0,1\}^\ell, \lambda\leq \ell,$ define as follows: $\Pi.\mathsf{enc}(k,m) = m \oplus H(k)$
$\Pi.\mathsf{dec}: \mathcal{K} \times \{0,1\}^\ell \rightarrow \{0,1\}^\lambda$ define as follows: $\Pi.\mathsf{dec}(k,c) = c \oplus H(k)$

is $\Pi$ chosen plain-text secure (CPA)? To the best of my knowledge, if $H$ is modeled as a pseudorandom generator (PRG) or a pseudorandom function (PRF), then $\Pi$ is CPA if a distinguisher cannot distinguish the output of the PRG/PRP from an output produced by a truly random function. However, I am not able to find such a reduction for a CRH.
Let's say Alice uses $\Pi$ to share a message $m$ over pair-wise channels with Bob and Charles (Alice-Bob, Alice-Charles share a symmetric key produced by a Diffie-Hellman exchange).

Alice sends $c_b = m\oplus H(k_{AB})$ to Bob and $c_c = m\oplus H(k_{AC})$ to Charles

Let's assume Decisional Diffie-Hellman. if an adversary is able to break the CRH, can it extracts $m$ from $\{c_b, c_c\}$? (we can assume that besides Bob and Charles, Alice sends the same message to other people).
Edit: I think the answers to this question help explain why CRH is not enough to provide CPA security. However, in my opinion, it does not fully answer this question because, as mentioned in the answers below, simply providing a uniformly distributed output does not imply CPA. Non-determinism is needed.
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.