Yes. My favourite way to show that there is an independent family of size $\mathfrak{c}$ is to note that for any countable transitive model $M$ of enough of ZFC, there is a perfect set of mutual Cohen generics over $M$. Such a perfect set can be forced over $M$ in several ways. Here is one of them: 

Consider the partial order $\mathbb{P}$ consisting of monotone functions $h \colon 2^{\leq n} \to 2^{<\omega}$, for some $n \in \omega$, ordered by extension. If $G$ is $\mathbb{P}$-generic over $M$ and $H := \bigcup_{h \in G} h$, then $f \colon 2^\omega \to 2^\omega$ such that $f(x) = \bigcup_{n \in \omega} H(x\restriction n)$ is a continuous bijection between $2^\omega$ and $f''2^\omega$. 

By genericity it is easy to see that $f'' 2^\omega$ is independent. Moreover another genericity argument shows that for any infinite coinfinite $x \in 2^\omega$ (where we identify subsets of $\omega$ with their characteristic functions), $x$ hits and avoids $f(x)$ infinitely. Now just let $\mathcal{F} = f''\Omega$.