Claim.$\newcommand{\mc}[1]{\mathcal{#1}}\newcommand{\sm}{\setminus}$ Any infinite almost disjoint family $\mc A$ is uncountable.
Proof. Let $\mc A=\{A_n; n<\omega\}$ be a countable AD-family. By induction we define \begin{align*} b_1&=\min A_1\\ b_2&=\min \{k\in A_2\sm A_1; k>b_1\}\\ b_n&=\min \{k\in A_n\sm\bigcup\limits_{j=0}^{n-1}A_j; k>b_n\}\\ \end{align*}
The set $B=\{b_n; n\in\omega\}$ is an infinite set which has a finite intersection with every set from$\mc A$. (Just notice that $B\cap A_n\subseteq \{b_1,b_2,\dots,b_n\})$. Therefore $\mc A$ is not maximal the set $B$ can be added.
Corollary. Any MAD-family is uncountable.
This basically says that for the almost disjoint number $\mathfrak a$ (defined as the smallest cardinality of a MAD-family on $\omega$) we have $\aleph_1\le \mathfrak a\le\mathfrak c$.
Corollary. There is no MAD-family $\mc M$ with the property described in the question.
Proof. If we had $|R\cap M|=1$ for each $M\in \mc M$, this would give us a bijection between $R$ and $\mc M$. So $\mc M$ would be countable, a contradiction.