I believe the following provides a counter-example for "weak implies norm continuity" in the case $E=c_0$.
Given any $F:\overline{\mathbb D} \rightarrow c_0$ let $F_n:\overline{\mathbb D} \rightarrow \mathbb C$ be the $n$th coordinate. If $F$ is weakly continuous on $\overline{\mathbb D}$ and analytic on $\mathbb D$, then each $F_n$ is a member of $A(\mathbb D)$ and $\|F_n\|_\infty\leq K$ for some $K$ independent of $n$. Conversely, if these two conditions hold on the $F_n$ then for any $a=(a_n)\in\ell^1 = c_0^*$ we have that $$ \newcommand{\ip}[2]{\langle{#1},{#2}\rangle}\ip{a}{F(z)} = \sum_n a_n F_n(z) $$ and so $z\mapsto \ip{a}{F(z)}$ is in $A(\mathbb D)$ as it is the absolutely convergent sum of members of $A(\mathbb D)$. By properties of functions in Hardy spaces (see e.g. page 50 of these notes) we know that $$ F_n(\xi) = \int_{\mathbb T} \frac{z F_n(z)}{z-\xi} \ dz \qquad (\xi\in\mathbb D). $$
We now seek to construct $F$ by constructing a suitable sequence $(F_n)$. That $F(z)\in c_0$ for all $z$ implies that $F_n(z)\rightarrow 0$ pointwise. Under the assumption that $\|F_n\|_\infty\leq K$ for all $n$, for fixed $\xi\in\mathbb D$ we can apply the dominated convergence theorem to the integral above to see that $F_n(z)\rightarrow 0$ pointwise for $z\in\mathbb T$ implies that $F_n(\xi)\rightarrow 0$.
Finally, we shall construct $F_n$ as Outer functions (page 35 in the notes linked above). This means we can take the value of $F_n$ on $\mathbb T$ to be an arbitrary positive (continuous) function with $|\log F_n|$ integrable on $\mathbb T$; we also need $F_n(z)\leq K$ for all $z\in\mathbb T$ and $F_n(z)\rightarrow 0$ pointwise. Choose $t\mapsto F_n(e^{it})$ to be the piecewise linear function with $F_n(1) = F_n(e^{i2\pi/n})=1/n$ and $F_n(e^{i\pi/n})=1$.
Finally, we consider $$ \| F(e^{i\pi/n}) - F(1) \|_{\infty} \geq | F_n(e^{i\pi/n}) - F_n(1) | = 1 - 1/n $$ so that while $e^{i\pi/n}\rightarrow 1$ we have that $F(e^{i\pi/n})\not\rightarrow F(1)$ in $c_0$. So $F$ is not norm continuous.