At least if $I$ is countable this should be true, but I do not have a reference:

Let $H$ be the CM space of the Frechet space $\mathcal{B} := \prod_{\alpha \in I} \mathcal{B}_{\alpha}$ with the product topology and the product measure. Recall that the Hilbert space reproducing kernel of $(\mathcal{B}, \mu)$ is isometrically isomorphic ($\Vert \cdot \Vert_{L^2(\mathcal{B}^{\ast}, \mu)} = \Vert \cdot \Vert_H$) to the CM space $H$ and is defined as the $L^2$ closure of $j(\mathcal{B}^{\ast})$ where $j$ is the canonical inclusion $\mathcal{B}^{\ast} \rightarrow L^2(\mathcal{B}, \mu)$.

In our case, the dual to $\mathcal{B}$ can be identified with the sequences in $\prod_{\alpha \in I} \mathcal{B}_{\alpha}^{\ast}$ with finite support and the pairing $\mathcal{B}^{\ast} \times \mathcal{B} \rightarrow \mathbb{R}$ given by

\begin{equation}
v(x) = \sum_{\alpha \in I} v_{\alpha} (x_{\alpha}), ~~ v_{\alpha} \in \mathcal{B}_{\alpha}^{\ast}, x_{\alpha} \in \mathcal{B}_{\alpha} .
\end{equation}

where the sum is, of course, finite.

Restricted to $j(\mathcal{B}^{\ast})$, the $L^2$-norm is

\begin{equation}
\Vert v \Vert_H = \Vert j(v) \Vert_{L^2} = \int \sum_{\alpha, \beta \in I} v_{\alpha} (x_{\alpha}) v_{\beta} (x_{\beta}) d \mu(x) = \sum_{\alpha \in I} \int v_{\alpha} (x_{\alpha})^2 d \mu(x) = \sum_{\alpha \in I} \Vert v_{\alpha} \Vert_{H_{\alpha}}^2
\end{equation}

where we used the product structure of $\mu$. Hence indeed

\begin{equation}
H = \{\mathbf v=\{v_\alpha\}_{\alpha\in I}:\sum_{\alpha\in I}\|v_\alpha\|_{\mu_\alpha}^2<\infty\} .
\end{equation}