This answer only deals with the case that $R$ is infinite. I thought that I would be able to modify it to the finite case - thanks to Ilya Bogdanov for spotting the mistake in my argument. (His answer shows that for finite $R$ such family indeed exists.) And thanks to bof for explaining in a comment that my original argument was unnecessarily complicated.
$\newcommand{\mc}[1]{\mathcal{#1}}$Let us assume that $R$ and $\mc M$ fulfill the conditions given in the question. Moreover, let $R$ be an infinite set.
Then the system $\{R\}\cup\mc M$ is an almost disjoint family. (For every $M\in \mc M$, the intersection $R\cap M$ is finite.) Thus from maximality we get $R\in\mc M$.
But now $|R\cap R|=|R|\ne 1$, contradicting "choosability".
The above argument can be shortly summarized as follows: If an infinite sets has a finite intersection with each element of a MAD family $\mc M$, then this set belongs to $\mc M$.