The answer is yes. Hugh Woodin showed me the following argument, which I post here with his permission.

Let $\varphi(v)$ be a formula in the language of set theory expanded by a unary predicate symbol. Given a pair of ordinals $(\alpha, \beta)$, working in $V^{\text{Col}(\omega,\alpha)}$ we let $B$ be a universally Baire set of reals having Wadge rank $\beta$ in the model $L(B,\mathbb{R})$, which satisfies $\mathsf{AD}^+$. Note that this model depends only on $\beta$ and not on $B$, and also that every set of reals in $L(B,\mathbb{R})$ is universally Baire because $B^\sharp$ exists and is universally Baire. Let $T_{\alpha,\beta}$ be the tree of a $(\Sigma^2_1)^{L(B,\mathbb{R})}$-scale on the set $$\{x \in \mathbb{R} : \exists C \in L(B,\mathbb{R})\, (\text{HC}; \in, C) \models \varphi[x]\}.$$ By the homogeneity of $\text{Col}(\omega,\alpha)$ this tree is is independent of the choice of generic filter and we have $T_{\alpha,\beta} \in V$. Let $T$ by the amalgamation of all the trees $T_{\alpha,\beta}$, so that $T$ is a tree on $\omega \times \text{Ord}$ and $p[T] = \bigcup_{\alpha,\beta \in \text{Ord}} p[T_{\alpha,\beta}]$ in every generic extension of $V$.

We claim that $$ V^{\text{Col}(\omega,\alpha)} \models p[T] = \{x \in \mathbb{R} : \exists C \in \text{uB}\, (\text{HC}; \in, C) \models \varphi[x]\},$$ for every ordinal $\alpha$.

From the definition of the trees $T_{\alpha,\beta}$ it is easy to see that $$ V^{\text{Col}(\omega,\alpha)} \models \bigcup_{\beta \in \text{Ord}} p[T_{\alpha,\beta}] = \{x \in \mathbb{R} : \exists C \in \text{uB}\, (\text{HC}; \in, C) \models \varphi[x]\},$$ so it remains to prove the left-to-right inclusion in the claim. Let $G \subset \text{Col}(\omega,\alpha)$ be a $V$-generic filter and let $x \in p[T]^{V[G]}$, say $x \in p[T_{\alpha',\beta'}]$ for ordinals $\alpha'$ and $\beta'$. We want to show
\begin{equation*}\tag{$*$} \exists C \in \text{uB}^{V[G]}\, (\text{HC}^{V[G]}; \in, C) \models \varphi[x]. \end{equation*}

If $\alpha' = \alpha$, this is easy. There are two remaining cases to consider:

1. $\alpha' > \alpha$.

2. $\alpha' < \alpha$.

In case (1), we have ($*$) by $(\Sigma^1_2)^{\text{uB}}$ generic absoluteness for $\text{Col}(\omega,\alpha')$. In case (2), we use the fact that if $B \in V[G \restriction \alpha']$ is a universally Baire set as in the definition of the tree $T_{\alpha', \beta'}$, then $B^\sharp$ exists and is universally Baire, so there is an elementary embedding $$ j : L(B, \mathbb{R}^{V[G \restriction \alpha]}) \to L(B^{V[G]}, \mathbb{R}^{V[G]}),$$ and we have $j(T_{\alpha', \beta'}) = T_{\alpha, \beta}$ where $\beta$ is the Wadge rank of $B^{V[G]}$. Considering the pointwise image of a branch witnessing $x \in p[T_{\alpha',\beta'}]$, we have $x \in p[T_{\alpha,\beta}]$ . Therefore ($*$) is witnessed by a set of reals $C \in L(B^{V[G]}, \mathbb{R}^{V[G]})$.

The answer is yes. Hugh Woodin showed me the following argument, which I post here with his permission.

Let $\varphi(v)$ be a formula in the language of set theory expanded by a unary predicate symbol. Given a pair of ordinals $(\alpha, \beta)$, working in $V^{\text{Col}(\omega,\alpha)}$ we let $B$ be a universally Baire set of reals having Wadge rank $\beta$ in the model $L(B,\mathbb{R})$, which satisfies $\mathsf{AD}^+$. Note that this model depends only on $\beta$ and not on $B$, and also that every set of reals in $L(B,\mathbb{R})$ is universally Baire because $B^\sharp$ exists and is universally Baire. Let $T_{\alpha,\beta}$ be the tree of a $(\Sigma^2_1)^{L(B,\mathbb{R})}$-scale on the set $$\{x \in \mathbb{R} : \exists C \in L(B,\mathbb{R})\, (\text{HC}; \in, C) \models \varphi[x]\}.$$ By the homogeneity of $\text{Col}(\omega,\alpha)$ this tree is is independent of the choice of generic filter and we have $T_{\alpha,\beta} \in V$. Let $T$ by the amalgamation of all the trees $T_{\alpha,\beta}$, so that $T$ is a tree on $\omega \times \text{Ord}$ and $p[T] = \bigcup_{\alpha,\beta \in \text{Ord}} p[T_{\alpha,\beta}]$ in every generic extension of $V$.

We claim that $$ V^{\text{Col}(\omega,\alpha)} \models p[T] = \{x \in \mathbb{R} : \exists C \in \text{uB}\, (\text{HC}; \in, C) \models \varphi[x]\},$$ for every ordinal $\alpha$. The right-to-left inclusion follows immediately from the definition of the trees $T_{\alpha,\beta}$, so it remains to prove the left-to-right inclusion. Let $G \subset \text{Col}(\omega,\alpha)$ be a $V$-generic filter and let $x \in p[T]^{V[G]}$, say $x \in p[T_{\alpha',\beta'}]$ for ordinals $\alpha'$ and $\beta'$. We want to show
\begin{equation*}\tag{$*$} \exists C \in \text{uB}^{V[G]}\, (\text{HC}^{V[G]}; \in, C) \models \varphi[x]. \end{equation*}

If $\alpha' = \alpha$, this is easy. There are two remaining cases to consider:

1. $\alpha' > \alpha$.

2. $\alpha' < \alpha$.

In case (1), we have ($*$) by $(\Sigma^1_2)^{\text{uB}}$ generic absoluteness for $\text{Col}(\omega,\alpha')$. In case (2), we use the fact that if $B \in V[G \restriction \alpha']$ is a universally Baire set as in the definition of the tree $T_{\alpha', \beta'}$, then $B^\sharp$ exists and is universally Baire, so there is an elementary embedding $$ j : L(B, \mathbb{R}^{V[G \restriction \alpha]}) \to L(B^{V[G]}, \mathbb{R}^{V[G]}),$$ and we have $j(T_{\alpha', \beta'}) = T_{\alpha, \beta}$ where $\beta$ is the Wadge rank of $B^{V[G]}$. Considering the pointwise image of a branch witnessing $x \in p[T_{\alpha',\beta'}]$, we have $x \in p[T_{\alpha,\beta}]$ . Therefore ($*$) is witnessed by a set of reals $C \in L(B^{V[G]}, \mathbb{R}^{V[G]})$.