It is consistent that the answer is positive and it is consistent that the answer is negative. 

**Claim:** There is a generic extension, $V[G]$ by a weakly homogeneous forcing notion in which there is a rigid forcing notion $\mathbb{P}$ such that for every generic filter $H \subseteq \mathbb{P}$, $$HOD^V = HOD^{V[G]} = HOD^{V[G][H]}.$$

**Proof:** Let $\kappa$ be a regular uncountable cardinal. Recall that $\square(\kappa)$ sequence is a sequence $\mathcal{C} = \langle C_\alpha \mid \alpha < \kappa\rangle$ such that $C_\alpha \subseteq \alpha$ is a club, for every accumulation point $\beta \in \text{acc }C_\alpha$, $C_\beta = C_\alpha \cap \beta$ and there is no club $D \subseteq \kappa$ such that for every $\beta \in \text{acc }D$, $D \cap \beta = C_\beta$. 

Let $\mathbb{S}$ be the forcing notion for adding a $\square(\kappa)$ sequence using bounded approximations (or successor ordinal length). Let $\mathcal{C}$ be the generic $\square(\kappa)$ sequence. Let $\mathbb{P}$ be the forcing notion $\mathcal{C}$, where $C_\alpha$ is stronger than $C_\beta$ in the order of $\mathbb{P}$ if $C_\alpha$ is an end extension of $C_\beta$.

$\mathbb{S}$ is weakly homogeneous and $\mathbb{S} \ast \mathbb{P}$ has a dense subset isomorphic to the Cohen forcing $\text{Add}(\kappa, 1)$. Therefore the Boolean completion of $\mathbb{S} \ast \mathbb{P}$ is weakly homogeneous. On the other hand $\mathbb{P}$ is rigid. In fact, if $V[G] \subseteq W$ is a model of $ZFC$ and $H_1, H_2\in W$ are two distinct $V[G]$-generic filters for $\mathbb{P}$ then $W \models \text{cf }\kappa = \omega$. 

**Claim:** Assume $V = L$. Then for every complete Boolean algebra $\mathbb{B}$, $\Vdash_{\mathbb{B}} HOD^{V[G]} = L$ iff for densely many $b\in \mathbb{B}$, $\mathbb{B} \restriction b$ is weakly homogeneous.

**Proof:** Let $$\mathbb{C} = \{ x \in \mathbb{B} \mid \forall \sigma \in \text{Aut}({\mathbb{B}}),\ \sigma(x) = x\}.$$ 

Where we take the set of all automorphisms that respect also arbitrary $\inf$ and $\sup$.

$\mathbb{C}$ is a complete subalgebra of $\mathbb{B}$. Let $G \subseteq \mathbb{B}$ be $L$-generic and let us claim that the generic filter for $\mathbb{C}$ belongs to $HOD^{V[G]}$.

Indeed, $b \in G\cap \mathbb{C}$ iff for every $L$-generic filter $H\subseteq \mathbb{B}$, $b\in H$. This follows from a theorem of Vopenka and Hajek:

**Theorem (Vopenka and Hajek):** If $H_1, H_2\subseteq \mathbb{B}$ are $V$-generic and $V[H_1] = V[H_2]$ then there is an automotphism $\sigma\in \text{Aut}(\mathbb{B})$ such that $\sigma '' H_1 = H_2$. 

Thus, we conclude that $\Vdash \dot{G} \cap \mathbb{C}\in L$, and in particular, $\mathbb{C}$ is atomic. Below every atom of $\mathbb{C}$ the forcing $\mathbb{B}$ is weakly homogeneous, as the orbit of every element is a pre-dense set.