This is a partial answer. I will show that if $\delta$ is Woodin then $\diamondsuit_\delta$ holds.
Claim: Any Woodin cardinal is subtle.
Proof: Let $\delta$ be a Woodin cardinal. Let $\vec{A} = \langle A_\alpha \mid \alpha < \delta\rangle$ be a sequence a sets, $A_\alpha \subseteq \alpha$ and let $C$ be a club in $\delta$. We want to find $\alpha < \beta$ in $C$ such that $A_\alpha = A_\beta \cap \alpha$.
Since $\delta$ is Woodin, there is a cardinal $\kappa < \delta$ which is $\vec{A} \times C$-strong up to $\delta$. Thus, $\kappa \in C$ and there is an elementary emebedding $j\colon V\to M$, such that :
- $\mathrm{crit}\ j = \kappa$,
- $j(\vec{A}) \restriction \kappa + 1 = \vec{A} \restriction \kappa + 1$,
In $M$, $j(\vec{A})(j(\kappa)) \cap \kappa = j(\vec{A})(\kappa) = A_\kappa$ and $\kappa, j(\kappa) \in j(C)$. By elementarity, there is $\alpha < \kappa$ in $C$ such that $A_\alpha = A_\kappa \cap \alpha$.