Edit 2018-08-08: Partial positive result added (Theorem 2).
First, here is a counterexample for the case of weak${}^*$-continuity:
Counterexample 1. Let $E = \ell^\infty := \ell^\infty(\mathbb{N}_0)$ and let $f: \overline{\mathbb{D}} \to E$ be given by
\begin{align*}
z \mapsto f(z) := (z^n)_{n \in \mathbb{N_0}}.
\end{align*}
For every $\mu \in E_* = \ell^1$ the function
\begin{align*}
\mu \circ f: z \mapsto \sum_{n=0}^\infty \mu_n z^n
\end{align*}
is in $A(\mathbb{D})$, i.e. $f$ is weakly${}^*$-continuous on $\overline{\mathbb{D}}$ and weakly${}^*$-holomorphic on $\mathbb{D}$. Hence, $f$ is actually holomorphic on $\mathbb{D}$ (since $f$ is bounded and $E_*$ is norming in $E^*$ or, more generally, since $f$ is bounded and $E_* \subseteq E^*$ separates $E$; see e.g. [Arendt, Nikolski: Vector-valued holomorphic functions revisisted (2000), Theorems 1.3 and 3.1]).
However, $f$ is not norm continuous at $z=1$ as we have $\|f(1) - f(z)\|_\infty \ge 1$ for every $z \in \mathbb{D}$.
Now, let us prove a positive (no pun intended) result.
Theorem 2. Let $E$ be a complex Banach lattice and let $f: \overline{\mathbb{D}} \to E$ be a function which is holomorphic on $\mathbb{D}$ and weakly continuous on $\overline{\mathbb{D}}$. If $f^{(k)}(0) \ge 0$ for each $k \in \mathbb{N}_0$, then $f$ is continuous on $\overline{\mathbb{D}}$.
Proof. We set $a_k := f^{(k)}(0)/k!$ for each $k \in \mathbb{N}_0$. For every $0 \le \mu \in E^*$ we have
\begin{align*}
\sum_{k=0}^\infty \langle \mu, a_k\rangle = \lim_{r \uparrow 1} \sum_{k=0}^\infty r^k \langle \mu, a_k\rangle = \lim_{r \uparrow 1} \langle \mu, \sum_{k=0}^\infty a_k r^k \rangle = \lim_{r \uparrow 1} \langle \mu, f(r)\rangle = \langle \mu, f(1) \rangle,
\end{align*}
where the first equality follows from the monotone convergence theorem. Hence, the increasing sequence $(\sum_{k=0}^n a_k)_{n \in \mathbb{N}_0}$ converges weakly to $f(1)$, and thus the sequence is even norm convergent to $f(1)$ by the Banach lattice version of Dini's theorem (see e.g. [Schaefer: Banach Lattices and Positive Operators (1974), Corollary to Theorem II.5.9]).
Moreover, the net $(f(r))_{r \in [0,1)}$ is increasing and weakly convergent to $f(1)$, so it is in fact norm convergent to $f(1)$, again by Dini's theorem.
The convergence of the sequence $(\sum_{k=0}^n a_k)_{n \in \mathbb{N}_0}$ implies that, for every complex number $\lambda$ of modulus $1$, the sequence $(\sum_{k=0}^n \lambda^k a_k )_{n \in \mathbb{N}_0}$ is a Cauchy-sequence (and thus convergent) for we have
\begin{align*}
\lvert \sum_{k=n_1+1}^{n_2} \lambda^k a_k \rvert \le \sum_{k=n_1+1}^{n_2} a_k
\end{align*}
whenever $0 \le n_1 \le n_2$.
Now we show that the set $f(\mathbb{D})$ is relatively compact in $E$. To this end, let $\varepsilon > 0$ and choose $r_0 \in (0,1)$ such that $\lVert f(1) - f(r)\rVert < \varepsilon$ whenever $r \ge r_0$. We show that every vector in $f(\mathbb{D})$ is closer than $2\varepsilon$ to the compact set $f(r_0 \overline{\mathbb{D}})$
Let $\lambda$ be a complex number of modulus $1$ and let $r \in [r_0,1)$. Then we have
\begin{align*}
\lvert \sum_{k=0}^\infty \lambda^k a_k - f(r\lambda) \rvert \le \sum_{k=0}^\infty (1 - r^k)a_k = f(1) - f(r)
\end{align*}
and thus
\begin{align*}
\lVert \sum_{k=0}^\infty \lambda^k a_k - f(r\lambda) \rVert \le \lVert f(1) - f(r) \rVert < \varepsilon.
\end{align*}
This implies that $\lVert f(r\lambda ) - f(r_0 \lambda) \rVert < 2\varepsilon$ for every $r \in [r_0,1)$ and every complex number $\lambda$ of modulus $1$.
Thus, every vector in $f(\mathbb{D})$ is indeed closer than $2\varepsilon$ to a vector in the compact set $f(r_0 \overline{\mathbb{D}})$, as claimed. This proves that $f(\mathbb{D})$ is totally bounded and thus relatively compact.
Now, let $z_0 \in \partial \mathbb{D}$. First consider a sequence $(z_n)$ in $\mathbb{D}$ which converges to $z_0$. Then every subsequence of $(f(z_n))$ has a norm convergent subsequence, and the limit of this latter subsequence equals $f(z_0)$ since $(f(z_n))$ converges weakly to $f(z_0)$ by assumption.
Hence, we conclude that $(f(z_n))$ is actually norm convergent to $f(z_0)$.
A simple approximation argument now shows that we also have $\lim_{n \to \infty} f(z_n) = f(z_0)$ for every sequence in $\overline{\mathbb{D}}$ which converges to $z_0$. The theorem is proved.
Remark 3. Theorem 2 shows in particular that the function $f$ from Counterexample 1 cannot be weakly continuous on $\overline{\mathbb{D}}$.
