No, a given MAD family does not necessarily admit a maximal independent subset.

The key notion here is that of a *completely separable* MAD family. This means a MAD family $\mathcal A$ having the property that for every $X \subseteq \omega$, either $X \subseteq \bigcup \mathcal A_0$ for some finite $\mathcal A_0 \subseteq \mathcal A$, or else there is some $A \in \mathcal A$ with $A \subseteq X$.

It is consistent that such families exist.
It is still (I think) an open question whether $\mathsf{ZFC}$ proves the existence of a completely separable MAD family. The question of whether this is true was raised by Erdős and Shelah in the early 70's. Balcar and Simon proved that it is consistent for completely separable MAD families to exist under a wide variety of hypotheses ($\mathfrak{a} = \mathfrak{c}$, or $\mathfrak{b} = \mathfrak{d}$, or $\mathfrak{d} \leq \mathfrak{a}$, or $\mathfrak{s} = \aleph_1$). Years later, Shelah proved that the existence of completely separable MAD families follows from either $\mathfrak{s} \leq \mathfrak{a}$, or else $\mathfrak{a} < \mathfrak{s}$ plus a certain PCF hypothesis that might be (but is not known to be) a theorem of $\mathsf{ZFC}$. The failure of this hypothesis implies $\mathfrak{c} > \aleph_\omega$ and (I think) the consistency of large cardinals. More information can be found in this paper of Osvaldo Guzman (see especially the bottom of page 2 and top of page 3).

This is relevant to your question because:

**Observation:** If $\mathcal A$ is a completely separable MAD family (and $\mathcal A$ is infinite), then there is no maximal independent set for $\mathcal A$.

The reason: If $I \subseteq \omega$ and $I \not\supseteq A$ for every $A \in \mathcal A$, then because $\mathcal A$ is completely separable, there is some finite $\mathcal A_0 \subseteq \mathcal A$ with $I \subseteq \bigcup \mathcal A_0$. But (as $\mathcal A$ is infinite) $\omega \setminus \bigcup \mathcal A_0$ is infinite. But if $n \in \omega \setminus \bigcup \mathcal A_0$, then $I \cup \{n\}$ is still "independent" in the sense that $I \cup \{n\} \not\supseteq A$ for every $A \in \mathcal A$. Thus $I$ is not maximal with respect to this property.