We suppose there is a proper class of Woodin cardinals. We claim that the statement $\phi^*$ which says that ``$\exists \mathbb{B} \exists\alpha(\mathbb{B}\text{ is a complete Boolean algebra and } \alpha\text{ is an ordinal}\wedge V_{\alpha}^{\mathbb{B}}\models \phi)"$, where $\phi$ is a $\Sigma_2$ sentence is absolute between $V$ and $V^{\mathbb{H}}$ for any complete Boolean algebra $\mathbb{H}$.

Notice that $\phi^*$ is a $\Sigma_2$ sentence since the statement $``V^{\mathbb{B}}\models \phi"$ is $\Delta_2$ and the statement $``\mathbb{B}$ is a complete Boolean algebra" is just $\Pi_1$. Let then $\phi$ be any $\Sigma_2$-sentence.

First, let $\mathbb{H}$ be any complete Boolean algebra. Let $\check{\gamma}\in V^{\mathbb{H}}$. We want to find a complete Boolean algebra $\mathbb{B}\in V^{\mathbb{H}}$ such that we have $$V^{\mathbb{H}}\models ``V_{\check{\gamma}}^{\mathbb{B}}\models \phi".$$ But since the forcing extension of a $V_{\gamma}$ is the same as the $V_{\gamma}$ of a forcing extension, that is $V_{\gamma}^{\mathbb{H}}=(V_{\check{\gamma}})^{V^{\mathbb{H}}}$. Since there is a complete Boolean algebra $\mathbb{B}$ in $V$ such that $\phi$ holds in the extension $V^{\mathbb{B}}$, one may then find a complete Boolean algebra $\mathbb{B}\in V^{\mathbb{H}}$ such that we can conclude that $$V^{\mathbb{H}}\models ``V^{\mathbb{B}}_{\check{\gamma}}\models \phi".$$

Next assume that for any complete Boolean algebra $\mathbb{H}$, the statement $\phi^*$ hold in $V^{\mathbb{H}}$ for some $\Sigma_2$ sentence $\phi$. Let $\mathbb{H}$ have cardinality $\gamma$. By assumption we let $\gamma<\delta$ be a Woodin cardinal. The forcing notion which will force $\phi$ over $V$ is given by the stationary tower forcing. Define the set

$$Y=\{X: X\prec H(\gamma^+)\wedge \vert X\vert =\aleph_0\}.$$

Define the stationary tower forcing to be the forcing $$\mathbb{Q}_{<\delta}=\{X: X\text{ is stationary and }X\in V_{\delta}\}.$$ The order is defined by $X\leq Y$ if and only if $\bigcup Y\subseteq \bigcup X$ and $\{Z\cap (\bigcup Y): Z\in X\}\subseteq Y$. We must have that $Y\in \mathbb{Q}_{<\delta}$ since if $F: H(\gamma^+)^{<\omega}\to H(\gamma^+)$ and building the elementary substructure of $H(\gamma^+)$ closed under $F$, let's call it $X$, we have $$X\prec H(\gamma^+)$$ and $$\vert X\vert =\aleph_0$$ and $F"X\subseteq X$. This is exactly what's required by stationarity. Let $G\subseteq \mathbb{Q}_{<\delta}^X$ be $V$-generic where $\mathbb{Q}_{<\delta}^X$ is the stationary tower forcing restricted to $X$. BY the properties of the stationary tower forcing we can find in $V[G]$ an elementary embedding $$j: V[G] \to M$$ where $M$ is a transitive model of $\text{ZFC}$ and such that, given that $\delta$ is a Woodin cardinal $$M^{<\delta}\subseteq M$$ and $H(\kappa^+)$ is countable in $V[G]$. Thus we have that $\mathbb{H} \in M$. We may then find a generic filter in $M$ for $\mathbb{H}$. Call this generic $H$ and we thus have $$V[H]\subseteq V[G].$$ Looking at the quotient forcing, choose a forcing notion $\mathbb{T}$ and a generic $K$ for $\mathbb{T}$ such that $$V[H][K]=V[G]$$ Finally we conclude that $V[G]\models \phi$ because $V[G]=V[H][K]$ and $V[H][K]\models \phi$ by assumption since $M\models \phi$