Edit: My previous answer had a mistake. Here is an updated answer which still contains some hopefully useful information:
Assuming $\operatorname{cov}(\mathcal{M}) = \mathfrak{d}$, there is a positive answer.
Recall that $\mathfrak{d}= \kappa$ is least such that there is a sequence $\langle K_\alpha : \alpha < \kappa\rangle$ of compact subsets of $2^\omega$ such that $\bigcup_{\alpha < \kappa} K_\alpha = \Omega$.
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 $f''2^\omega$.
We will recursively construct continuous functions $f_\alpha \colon K_\alpha \to \Omega$ for $\alpha < \kappa$. Namely, given $\langle f_\beta: \beta < \alpha \rangle$ for $\alpha < \kappa$, let $M_\alpha$ be an elementary submodel of some large $H(\theta)$ of size $\vert \alpha \vert < \kappa$ containing all $f_\beta$ for $\beta < \alpha$ and $K_\alpha$. Force with $\mathbb{P}$ over $M_\alpha$ and get a generic continuous function $f \colon 2^\omega \to 2^\omega$. This is possible since $\mathbb{P}$ is a countable poset and $\alpha < \operatorname{cov}(\mathcal{M})$. Let $f_\alpha := f \restriction K_\alpha$.
Finally, let $F \colon \Omega \to \Omega$ be such that $F(x) = f_\alpha(x)$ for $\alpha$ least such that $x \in K_\alpha$. To see that $F''\Omega$ is independent we note that:
Claim Let $c$ be Cohen over $M$ and $B \in M$ an analytic independent family. Then $B \cup \{ c \}$ is independent.
Proof. Consider the set $A_0$ of $x \in \Omega$ such that for some Boolean combination $y$ of elements of $B$, $x \subset^* y$ and the set $A_1$ of $x$ such that for some Boolean combination $y$, $x \cap y$ is finite. By an argument of Miller (see ), both $A_0$ and $A_1$ must be meager. These are analytic sets which are in $M$. As $c$ is Cohen over $M$, it is neither a member of $A_0$ nor $A_1$, proving the claim.
Since $f_\alpha(x_0), \dots, f_\alpha(x_k)$ are mutually Cohen generic over $M_\alpha$, any Boolean combination is Cohen as well and can be added to any finite union of the compact sets $f_\beta ''K_\beta \in M$, for $\beta < \alpha$, to form an independent set. Thus also the union of all $f_\beta '' K_\beta$, $\beta \leq \alpha$ is independent.
Finally, to see that $x \in K_\alpha$ hits and avoids $f_\alpha(x)$ infinitely, make the following genericity argument: Let $h \in \mathbb{P}$ be arbitrary, say the domain of $h$ is $2^{\leq n}$ and the range is contained in $2^{\leq m}$. By compactness, there is $k > m$ such that for any $x \in K_\alpha$ $x(i) = 1$ for some $i \in (m,k]$. We can simply extend $h$ to $h'$ of domain $2^{\leq k}$ such that $h'(s)$ hits $s$ above $m$ for every $s \in 2^{\leq k}$. Similarly for avoiding instead of hitting. Since $K_\alpha \in M_\alpha$, this genericity argument can be made over $M_\alpha$.