I think this should work: for a connected group $G$. The Lie algebra action extends, by multiplying by $i$$J$, to a complex Lie algebra action. By compactness of $M$, these vector fields are all complete. So some covering group of the complexification acts, for example the universal covering group $\tilde{G}_{\mathbb{C}}$ acts. Hochschild, The Structure of Complex Lie Groups, chapter XVII, §5, proves that if $G$ admits a faithful finite dimensional representation, then $G$ injects into its complexification. For example, all compact Lie groups admit faithful finite dimensional representations. The group $G$ is then the maximal compact subgroup of $G_{\mathbb{C}}$; Broecker and tom Dieck, Representations of Compact Lie Groups, p. 153, prop. 8.3. Every connected Lie group retracts to its maximal compact subgroup. In particular, the fundamental group of $G$ and of $G_{\mathbb{C}}$ are the same, represented by loops in $G$. The fundamental group of $G_{\mathbb{C}}$ embeds into the universal covering group, and we $\tilde{G}_{\mathbb{C}}$. We need to see which elements of it act trivially. So if, so that we can see which quotient of $\tilde{G}_{\mathbb{C}}$ the action descends to. The fundamental group of the complexification is the same as that of the original compact group sits as a subgroup of $\tilde{G}$, i.ebut acts trivially on $M$. ifSo the original compact groupsame is a maximal compact subgroup, then thattrue for the fundamental group of $G_{\mathbb{C}}$: $\pi_1(G_{\mathbb{C}}) \subset \tilde{G}_{\mathbb{C}}$ acts trivially on $M$. Hence the complexification$G_{\mathbb{C}}$ acts on the complex homogeneous space as long as the compact injects into the complexification$M$.
