Given an ultrafilter $\mathcal{U}$ on $\omega$ the corresponding Mathias forcing $\mathbb{M}_\mathcal{U}$ is the forcing consisting of conditions $\langle s,A\rangle$ where $s$ is a finite subset of $\omega$ and $A\in\mathcal{U}$. The ordering is given by $\langle t,B\rangle\leq\langle s,A\rangle$ if $B\subseteq A$ and $t$ is an end extension of $s$, with $t\setminus s\subseteq A$. (So $\mathbb{M} _\mathcal{U}$ is just like Prikry forcing, but the relevant objects are defined on $\omega$.)

An ultrafilter $\mathcal{U}$ is nowhere dense if whenever $F:\omega\rightarrow\mathbb{R}$ there is some $A\in\mathcal{U}$ whose image under $A$ is nowhere dense. Blaszczyk and Shelah have shown that when $\mathcal{U}$ is nowhere dense, then $\mathbb{M} _\mathcal{U}$ does not add a Cohen real. In fact, they proved that the existence of a $\sigma$-centered forcing adding no Cohen real is equivalent to the existence of a nowhere dense ultrafilter.

Their proof that $\mathbb{M} _\mathcal{U}$ adds no Cohen real mostly uses topological methods rather than forcing theoretic ones, and I find it a bit opaque. I was wondering if a simpler proof was known if we add some stronger hypotheses on the ultrafilter. My question is:

Is there a (relatively) simple proof that when $\mathcal{U}$ is a Ramsey ultrafilter $\mathbb{M}_\mathcal{U}$ adds no Cohen real?