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 $F_n(\xi)\rightarrow 0$.

We shall use Outer functions (page 35 in the notes linked above). Choose a positive function $h$ on $\mathbb T$ with $\log h$ being integrable; then we can form an outer function $F$ with $|F|$ having nontangential limit $h$ at the boundary, almost everywhere. It is not clear to me that just because $h$ is continuous it will follow that $F$ is in $A(\mathbb D)$ instead of just a Hardy space; so we work a bit harder.

Firstly, choose $t\mapsto h_n(e^{it})$ to be the piecewise linear function with $h_n(1) = h_n(e^{i2\pi/n})=1/n$ and $h_n(e^{i\pi/n})=1$. Let $F_n$ be the Outer function with $|F_n|$ having nontangential limit $h_n$ on the boundary, almost everywhere. We can find $r_n<1$ such that if we define $G_n(z) = F_n(r_nz)$ for $z\in\overline{\mathbb D}$ then there will be $a_n, b_n\in\mathbb T$ with $|a_n-1|<\frac1n, |b_n-e^{i\pi/n}|<\frac1n, ||G_n(a_n)|-\frac1n|<\frac1n$ and $||G_n(b_n)|-1|<\frac1n$. Certainly $G_n\in A(\mathbb D)$ with $\|G_n\|\leq 1$.

Finally, form $G\in A(\mathbb D, c_0)$ using the sequence $(G_n)$, and observe that $$ \| G(b_n) - G(a_n) \|_{\infty} \geq | G_n(b_n) - G_n(a_n) | \geq 1 - 3/n $$ so as $a_n \rightarrow 1$ and $b_n\rightarrow 1$, we conclude that $G$ cannot be norm continuous.

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 $F_n(\xi)\rightarrow 0$.

We shall use Outer functions (page 35 in the notes linked above). Choose a positive function $h$ on $\mathbb T$ with $\log h$ being integrable; then we can form an outer function $F$ with $|F|$ having nontangential limit $h$ at the boundary, almost everywhere. It is not clear to me that just because $h$ is continuous it will follow that $F$ is in $A(\mathbb D)$ instead of just a Hardy space; so we work a bit harder.

Firstly, choose $t\mapsto h_n(e^{it})$ to be the piecewise linear function with $h_n(1) = h_n(e^{i2\pi/n})=1/n$ and $h_n(e^{i\pi/n})=1$. Let $F_n$ be the Outer function with $|F_n|$ having nontangential limit $h_n$ on the boundary, almost everywhere. We can find $r_n<1$ such that if we define $G_n(z) = F_n(r_nz)$ for $z\in\overline{\mathbb D}$ then there will be $a_n, b_n\in\mathbb T$ with $|a_n-1|<\frac1n, |b_n-e^{i\pi/n}|<\frac1n, ||G_n(a_n)|-\frac1n|<\frac1n$ and $||G_n(b_n)|-1|<\frac1n$. Certainly $G_n\in A(\mathbb D)$ with $\|G_n\|\leq 1$.

Finally, form $G\in A(\mathbb D, c_0)$ using the sequence $(G_n)$, and observe that $$ \| G(b_n) - G(a_n) \|_{\infty} \geq | G_n(b_n) - G_n(a_n) | \geq 1 - 3/n $$ so as $a_n \rightarrow 1$ and $b_n\rightarrow 1$, we conclude that $G$ cannot be norm continuous.

Alternative construction (from a hint from Yemon Choi): Simply directly define $$ F_n(z) = \exp\big(k_n(e^{-i\pi/n}z-1)\big). $$ where $(k_n)$ is a rapidly increasing sequence. Then obviously $F_n$ is in the disc algebra. We have that $$ |F_n(e^{it})| = \exp\big( k_n(\cos(t-\pi/n)-1) \big). $$ For $t\not=0$ this obviously converges to $0$; we chose $(k_n)$ increasing fast enough so that $\exp( k_n(\cos(-\pi/n)-1) )\rightarrow 0$. Thus $F_n(z)\rightarrow 0$ pointwise on $\mathbb T$, and so on $\mathbb D$, as before. The remainder of the argument also runs the same way, as $F_n(1)\rightarrow 0$ while $F_n(e^{i\pi/n}) = 1$.

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.

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 $F_n(\xi)\rightarrow 0$.

We shall use Outer functions (page 35 in the notes linked above). Choose a positive function $h$ on $\mathbb T$ with $\log h$ being integrable; then we can form an outer function $F$ with $|F|$ having nontangential limit $h$ at the boundary, almost everywhere. It is not clear to me that just because $h$ is continuous it will follow that $F$ is in $A(\mathbb D)$ instead of just a Hardy space; so we work a bit harder.

Firstly, choose $t\mapsto h_n(e^{it})$ to be the piecewise linear function with $h_n(1) = h_n(e^{i2\pi/n})=1/n$ and $h_n(e^{i\pi/n})=1$. Let $F_n$ be the Outer function with $|F_n|$ having nontangential limit $h_n$ on the boundary, almost everywhere. We can find $r_n<1$ such that if we define $G_n(z) = F_n(r_nz)$ for $z\in\overline{\mathbb D}$ then there will be $a_n, b_n\in\mathbb T$ with $|a_n-1|<\frac1n, |b_n-e^{i\pi/n}|<\frac1n, ||G_n(a_n)|-\frac1n|<\frac1n$ and $||G_n(b_n)|-1|<\frac1n$. Certainly $G_n\in A(\mathbb D)$ with $\|G_n\|\leq 1$.

Finally, form $G\in A(\mathbb D, c_0)$ using the sequence $(G_n)$, and observe that $$ \| G(b_n) - G(a_n) \|_{\infty} \geq | G_n(b_n) - G_n(a_n) | \geq 1 - 3/n $$ so as $a_n \rightarrow 1$ and $b_n\rightarrow 1$, we conclude that $G$ cannot be norm continuous.