This is more similar to the result by Birman-Hilden that the classical braid group is a subgroup of the mapping class group of a surface with one or two boundary components (see Section 9.4 in Farb-Margalit's Primer), so I'm not sure if this answers your question.

Perron-Vannier prove that the Artin group $A(D_n)$ "geometrically embeds" into the mapping class group of a surface, i.e., that there is a surface $\Sigma$ with boundary (and no punctures) and a faithful (albeit not surjective) representation $A(D_n) \to \rm{Mod}(\Sigma)$ mapping the standard generators to Dehn twists.

The surface $\Sigma$ can be chosen as follows. Take a disc $\Delta$ and embedded arcs in $\Delta$ that intersect in the pattern of $D_n$ (edges in $D_n$ correspond to intersection points and vertices to arcs). Close up each arc with a band to form $\Sigma$. The Dehn twists along the closed arcs then generate a subgroup of $\rm{Mod}(\Sigma)$ isomorphic to $D_n$. In fact, the mentioned Dehn twists map to the standard generators of $A(D_n)$.

Labruère later studied the same construction of $\Sigma$, but taking the intersection pattern of a cycle $\widetilde A_{n-1}$ rather than $D_n$. The kernel of the resulting (still not surjective) representation $A(\widetilde A_{n-1}) \to \rm{Mod}(\Sigma)$ is generated by the so-called "cycle relation", and one can show that the quotient of $A(\widetilde A_{n-1})$ by that relation is again isomorphic to $A(D_n)$ (for example, using Baader-Lönne's results).