We can detect the difference between the two constructions using the homotopy category. Given any simplicially enriched category $\mathcal{C}$, we can construct an ordinary category $\pi_0 [\mathcal{C}]$ by applying $\pi_0$ to the hom-spaces. (This makes sense because $\pi_0 : \mathbf{sSet} \to \mathbf{Set}$ preserves finite products.) One might call the morphisms in $\mathcal{C}$ that become isomorphisms in $\pi_0 [\mathcal{C}]$ "(simplicial) homotopy equivalences".
Now, in general, $\mathcal{C}$ is not fibrant (i.e. a Kan-enriched category) – but we can fibrantly replace it by applying $\mathrm{Ex}^\infty : \mathbf{sSet} \to \mathbf{sSet}$ to its hom-spaces. Of course, the induced functor $\pi_0 [\mathcal{C}] \to \pi_0 [\mathrm{Ex}^\infty [\mathcal{C}]]$ is an isomorphism, and the homotopy coherent nerve of $\mathrm{Ex}^\infty [\mathcal{C}]$ is a quasicategory whose homotopy category is isomorphic to $\pi_0 [\mathcal{C}]$.
On the other hand, if $\mathcal{C}$ is the hammock localisation of a model category $\mathcal{M}$ (not necessarily simplicial), then $\pi_0 [\mathcal{C}]$ is isomorphic to $\operatorname{Ho} \mathcal{M}$. But even if $\mathcal{M}$ is a simplicial model category, $\pi_0 [\mathcal{M}]$ and $\operatorname{Ho} \mathcal{M}$ need not be isomorphic. (Exercise: Show this is the case for $\mathcal{M} = \mathbf{sSet}$ with the Kan–Quillen model structure.) Thus, $\mathcal{C}$ and $\mathcal{M}$ need not be weakly equivalent, so the homotopy coherent nerves of their fibrant replacements need not be equivalent. (Recall that the homotopy coherent nerve is a right Quillen equivalence!)
Finally, if $\mathcal{M}_\mathrm{cf}$ is the (simplicially enriched) full subcategory of cofibrant–fibrant objects in a simplicial model category $\mathcal{M}$, then $\mathcal{M}_\mathrm{cf}$ is fibrant, and as Aaron says, it is a theorem of Dwyer and Kan that $\mathcal{M}_\mathrm{cf}$ and $\mathcal{C}$ are weakly equivalent. Hence the homotopy coherent nerves of $\mathcal{M}_\mathrm{cf}$ and $\mathrm{Ex}^\infty [\mathcal{C}]$ are equivalent.