Neither such $f$ nor such $F$ exists.
Firstly, note that there exists a biholomorphic map
$$f_0:\Delta\to\Omega,\quad z\mapsto\left(\frac{1+z}{1-z}\right)^2. $$
Secondly, let us show that such $f$ does not exist. Given a biholomorphic map $f:\Delta\to\Omega$, $f_0^{-1}\circ f$ is an automorphism of the unit disk, i.e. $f=f_0\circ\varphi_{w,\theta}$ for some $w\in\Delta$ and $\theta\in\Bbb R$, where
$$\varphi_{w,\theta}(z)=e^{i\theta}\frac{z-w}{1-\bar{w}z}.$$
Note that both $f_0$ and $\varphi_{w,\theta}$ are holomorphic maps from $\widehat{\Bbb C}:=\Bbb C\cup\{\infty\}$ to itself, so $f$ is also a holomorphic map from $\widehat{\Bbb C}$ to itself. Then the assumption $\lim \limits_{z \to a} f(z) = \lim \limits_{z \to a} f(-z)$ when $|a| = 1$ is simply $f(a)=f(-a)$ when $|a| = 1$. Moreover, for $a\in\widehat{\Bbb C}$,
$$f(a)=\infty\iff a=\varphi_{w,\theta}^{-1}(1)\Rightarrow |a|=1.$$
For this $a$, $|a|=1$ and $f(a)=\infty$, but $-a\ne a$, so $f(-a)\ne\infty$, i.e. such $f$ does not exist.
Finally, let us show that such $F$ does not exist. Given a biholomorphic map $F:\Omega\to\Delta$, $f:=F^{-1}:\Delta\to\Omega$ is also biholomorphic. From the discussion on $f$ in the last paragraph we know that $f$ can be written as $f=f_0\circ\varphi_{w,\theta}$. Denote $g=\frac{1+\varphi_{w,\theta}}{1-\varphi_{w,\theta}}$. Then $f=g^2$ and hence $g^{-1}(z)=F(z^2)$, so when $x\le 0$,
$$\lim_{t\to 0^+}F(x\pm it)= g^{-1}(\pm i|x|^{\frac{1}{2}}).$$
As a result, if $F$ satisfies $\lim\limits_{t\to 0^+} F(x+it) = - \lim\limits_{t \to 0^+} F(x-it)$ when $x \le 0$, then $g$ satisfies $g^{-1}(-iy)=-g^{-1}(iy)$ when $y\in\Bbb R$. Letting $y\to\infty$ and noting that $g$ is a linear fractional transformation, we have
$$g^{-1}(\infty)=-g^{-1}(\infty)\Rightarrow g^{-1}(\infty)=0.$$
However,
$$g^{-1}(\infty)=\varphi_{w,\theta}^{-1}(1)\Rightarrow |g^{-1}(\infty)|=1,$$ a contradiction. Therefore, such $F$ does not exist.
