I have figured out how to prove a relatively weaker result.
By elementary inequality of matrix, we know that \begin{equation*} |X|_{\infty}\leq\|X\|\leq 1, \quad\text{ and }\quad\|X\|\leq \|X\|_F. \end{equation*} For every element $X_{i,j}$ of $X$, by Lemma 4.4.1 of this paper we have that \begin{equation*} \mathbb{E}\Big[\big|\mathbb{E}[X_{ij}|\mathcal{G}]\big|^2\Big]\leq \mathbb{E}\Big[\big|\mathbb{E}[X_{ij}|\mathcal{G}]\big|\Big]\leq 2\pi\phi. \end{equation*} Therefore, we have that \begin{align*} &\mathbb{E}\Big[\|\mathbb{E}[X|\mathcal{G}]\|\Big]\leq \mathbb{E}\Big[\|\mathbb{E}[X|\mathcal{G}]\|_F\Big],\\ \leq&\Big\{\mathbb{E}\Big[\|\mathbb{E}[X|\mathcal{G}]\|^2_F\Big]\Big\}^{1/2}\\ \leq&\Big\{\mathbb{E}\Big[\sum_{i,j=1}^d\big|\mathbb{E}[X_{ij}|\mathcal{G}]\big|^2\Big]\Big\}^{1/2}\leq\sqrt{d^22\pi\phi}=d\sqrt{2\pi\phi}. \end{align*}