Thanks everyone, for completeness I'll outline the proof here. The answer is **yes**, over $\mathbf{CMon}$, $\mathrm{DK}$ is fully faithful. To check this we define $N_*$ in the opposite direction taking $A$ to the complex $NA_n=\bigcap_{i\geq1}\ker{d_i}$ with differential $d_0$.

**Lemma** $\mathrm{DK}$ is left adjoint to $N_*$.

*Proof* We can view $\mathrm{Ch}_{\geq0}(\mathbf{CMon})$ as a presheaf category enriched in $\mathbf{CMon}$ over the category $Ch$ described in Tim's latest comment. Now it is clear $\mathrm{DK}$ preserves colimits since we can check it pointwise and it was defined using coproducts. So $\mathrm{DK}$ has a right adjoint, say $R$; note that the representables $r_n$ are the chain complexes of all zero and $\mathbb{N}\to\mathbb{N}$ as the $n$ and $n+1$ terms. Then $R$ satisfies
$$
R(A)_n=\mathrm{Hom}_{\mathbf{sCMon}}(\mathrm{DK}(r_n),A)\quad\text{ in }\mathbf{CMon}
$$
where we can see $\mathrm{DK}(r_n)_m$ is $0$ if $m<n$, $\mathbb{N}$ is $m=n$, and $\bigoplus_{i\geq1}^n\mathbb{N}$ if $m\geq n+1$. The definition of how $\mathrm{DK}(-)$ on face maps shows that we can identify this hom-monoid with $NA_n$, thus $\mathrm{DK}\dashv N_*$.

Now the result follows from the observation that $N_*\circ\mathrm{DK}\cong\mathrm{Id}$. Indeed, that $A_n\subset N_n\mathrm{DK}(A)_n$ is easy to see, and the reverse inclusion follows from the fact that for any surjection $\phi\colon[n]\to[k]$, we can find $i\geq1$ such that $\phi\circ d^i$ is still surjective (then the condition on kernels implies only the $A_n$ submonoid remains) [HA, Remark 1.2.3.11]. We are done as such an isomorphism implies this is a fully faithful adjoint pair.