I believe that I found the answer to this question. Let $ X $ be a Mori dream space which is of globally $ F $-regular type over a field $ L $ of characteristic zero. Assume that $ X $ is $ \mathbb{Q} $-factorial and normal. Because $ X $ is projective, there is a closed immersion $ \iota: X \to \mathbb{P}^{n}_{L} $. If $ L[W] $ is the homogeneous coordinate ring of $ \mathbb{P}^{n}_{L} $, then the ideal of $ X $ in $ \mathbb{P}^{n}_{L} $ is generated by homogeneous functions $ \{f_{i}(W)\}_{i=1}^{\ell} $. Each $ f_{i}(W) $ is equal to $ \sum_{I} a_{I,i} W^{I} $. There is an $ \mathbb{Z} $ algebra $ A $ such that $ A $ contains all non-zero $ a_{I,i} $ for $ 1 \le i \le \ell $.

Let $ \mathcal{X} $ be the sub-scheme of $ \mathbb{P}^{n}_{A} $ equal to $ \mathcal{Z}(\langle f_{i}(W) \rangle) $. The scheme $ \mathcal{X} $ is clearly flat over $ \operatorname{Spec}(A) $. Because the Cox sheaf of $ \mathcal{X} $ is flat as an $ \mathcal{O}_{\mathcal{X}} $-module, and $ \mathcal{X} $ is flat over $ \operatorname{Spec}(A) $, the scheme $ \operatorname{Spec}(H^{0}(\mathcal{X}, \mathcal{C} \operatorname{ox}(\mathcal{X}))) $, which we shall denote by $ \mathcal{Y} $ is flat over $ \operatorname{Spec}(A) $. Here, the sheaf $ \mathcal{C}\operatorname{ox}(\mathcal{X}) $ is the Cox sheaf of $ \mathcal{X} $. It is well defined because $ \mathcal{X} $ is flat over $ \operatorname{Spec}(A) $, which means that each of the fibres of $ \mathcal{X} $ over $ \operatorname{Spec}(A) $ are $ \mathbb{Q} $-factorial and normal. The generic fibre of $ \mathcal{Y} $ is isomorphic to $ \operatorname{Spec}(\operatorname{Cox}(X)) $.

The closed fibres of the scheme $ \mathcal{X} $ over $ \operatorname{Spec}(A) $ are globally $ F $-regular, by our assumption that $ X $ is of globally $ F $-regular type. Therefore, if $ a \in \operatorname{Spec}(A) $ is a closed point, then $ \operatorname{Cox}(\mathcal{X}_{a}) $ is $ F $-regular.

Because $ X $ is a Mori dream space, the Cox ring of $ X $ is finitely generated. If $ a \in \operatorname{Spec}(A) $ is a closed point, then since $ \mathcal{Y} $ is flat over $ \operatorname{Spec}(A) $, and each fibre $ \mathcal{Y}_{a} $ is isomorphic to $ \operatorname{Spec}(\operatorname{Cox}(\mathcal{X}_{a})) $, the scheme $ \operatorname{Spec}(\operatorname{Cox}(\mathcal{X}_{a})) $ is of finite type over $ k(a) $. An $ F $-finite, $ F $-regular ring is Cohen Macaulay. As a result if $ a \in \operatorname{Spec}(A) $ is a closed point, then each fibre $ \mathcal{Y}_{a} $ is Cohen Macaulay because $ \operatorname{Cox}(\mathcal{X}_{a}) $ is $ F $-finite and $ F $-regular. Because there is a Zariski dense sub-scheme $ U \subseteq \operatorname{Spec}(A) $ such that $ \mathcal{Y}_{a} $ is Cohen Macaulay, the fibre $ \mathcal{Y}_{L} $ is Cohen Macaulay. Therefore $ \operatorname{Cox}(X) $ is Cohen Macaulay.

This is my own question, so I will not vote for my answer unless enough people vote for this answer. If someone sees an error in this proof I will really appreciate hearing about it. Thank you.