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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.)

$\newcommand{\mc}[1]{\mathcal{#1}}$Let us assume that $R$ and $\mc M$ fulfill the conditions given in the question.

Let $$R=R_1\cup R_2$$ be any decomposition of $R$ into two disjoint infinite subsets. (Here we're using the assumption that $R$ is infinite.)

Clearly, we have $|R_1\cap M|\le 1$ for each $M\in\mc M$. Therefore the system $\mc M\cup\{R_1\}$ is almost disjoint and thus $R_1\in\mc M$ (from maximality).

But this means that the intersection $$R_1\cap R=R$$ is infinite — contradicting the "choosability".

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 $\mc M$.

$\newcommand{\mc}[1]{\mathcal{#1}}$Let us assume that $R$ and $\mc M$ fulfill the conditions given in the question.

Case 1 - if $R$ is infinite. Let $$R=R_1\cup R_2$$ be any decomposition of $R$ into two disjoint infinite subsets.

Clearly, we have $|R_1\cap M|\le 1$ for each $M\in\mc M$. Therefore the system $\mc M\cup\{R_1\}$ is almost disjoint and thus $R_1\in\mc M$ (from maximality).

But this means that the intersection $$R_1\cap R=R$$ is infinite — contradicting the "choosability".

Case 2 - if $R$ is finite and it has at least two elements. We can see relatively easily that $$\mathcal M'=\mathcal M\cup\{M\cup R; M\in\mathcal M\}$$ is almost disjoint. Maximality then implies that for each $M\in\mathcal M$ we have $$M\cup R\in\mathcal M.$$ Then we have $|(M\cup R)\cap R|\ge 2$, a contradiction.

Case 3 - if $R$ is a singleton. In this case the condition $|M\cap R|=1$ actually means $R\subseteq M$. So we get $\bigcap \mc M \ne \emptyset$, a contradiction.

Note. Originally I have posted only the first part - I have completely missed that I need to deal with a possibility that $R$ is finite, too. (I have realized this only after another answer was posted.)

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.)

$\newcommand{\mc}[1]{\mathcal{#1}}$Let us assume that $R$ and $\mc M$ fulfill the conditions given in the question.

Let $$R=R_1\cup R_2$$ be any decomposition of $R$ into two disjoint infinite subsets. (Here we're using the assumption that $R$ is infinite.)

Clearly, we have $|R_1\cap M|\le 1$ for each $M\in\mc M$. Therefore the system $\mc M\cup\{R_1\}$ is almost disjoint and thus $R_1\in\mc M$ (from maximality).

But this means that the intersection $$R_1\cap R=R$$ is infinite — contradicting the "choosability".

$\newcommand{\mc}[1]{\mathcal{#1}}$Let us assume that $R$ and $\mc M$ fulfill the conditions given in the question.

Case 1 - if $R$ is infinite. Let $$R=R_1\cup R_2$$ be any decomposition of $R$ into two disjoint infinite subsets.

Clearly, we have $|R_1\cap M|\le 1$ for each $M\in\mc M$. Therefore the system $\mc M\cup\{R_1\}$ is almost disjoint and thus $R_1\in\mc M$ (from maximality).

But this means that the intersection $$R_1\cap R=R$$ is infinite — contradicting the "choosability".

Case 2 - if $R$ is finite and it has at least two elements. We can see relatively easily that $$\mathcal M'=\mathcal M\cup\{M\cup R; M\in\mathcal M\}$$ is almost disjoint. Maximality then implies that for each $M\in\mathcal M$ we have $$M\cup R\in\mathcal M.$$ Then we have $|(M\cup R)\cap R|\ge 2$, a contradiction.

Case 3 - if $R$ is a singleton. In this case the condition $|M\cap R|=1$ actually means $R\subseteq M$. So we get $\bigcap \mc M \ne \emptyset$.

$\newcommand{\mc}[1]{\mathcal{#1}}$Let us assume that $R$ and $\mc M$ fulfill the conditions given in the question.

Case 1 - if $R$ is infinite. Let $$R=R_1\cup R_2$$ be any decomposition of $R$ into two disjoint infinite subsets.

Clearly, we have $|R_1\cap M|\le 1$ for each $M\in\mc M$. Therefore the system $\mc M\cup\{R_1\}$ is almost disjoint and thus $R_1\in\mc M$ (from maximality).

But this means that the intersection $$R_1\cap R=R$$ is infinite — contradicting the "choosability".

Case 2 - if $R$ is finite and it has at least two elements. We can see relatively easily that $$\mathcal M'=\mathcal M\cup\{M\cup R; M\in\mathcal M\}$$ is almost disjoint. Maximality then implies that for each $M\in\mathcal M$ we have $$M\cup R\in\mathcal M.$$ Then we have $|(M\cup R)\cap R|\ge 2$, a contradiction.

Case 3 - if $R$ is a singleton. In this case the condition $|M\cap R|=1$ actually means $R\subseteq M$. So we get $\bigcap \mc M \ne \emptyset$, a contradiction.

Note. Originally I have posted only the first part - I have completely missed that I need to deal with a possibility that $R$ is finite, too. (I have realized this only after another answer was posted.)