Put $T = G_r$. We may, and do, assume that $\beta = r$, and simply describe a $T$-invariant complex structure on $\operatorname T_{\mathcal O_r}(r)$.
Instead of having one 1-dimensional subspace of $\mathfrak g$ for every root $\alpha$, we get a $2$-dimensional subspace $\{X_\alpha + \overline{X_\alpha} \mathrel: X_\alpha \in (\mathfrak g_{\mathbb C})_\alpha\}$ for every pair of roots $\{\alpha, \overline\alpha = -\alpha\}$.
Our Kähler structure treats this $2$-dimensional space, which I will provocatively call $\mathfrak g_{\pm\alpha}$ since its complexification is $(\mathfrak g_{\mathbb C})_\alpha \oplus (\mathfrak g_{\mathbb C})_{-\alpha}$, as a $\mathbb C$-vector space via the (isomorphic) projection to $(\mathfrak g_{\mathbb C})_\alpha$, and then rotates by $i$—but we must choose $i$ appropriately to get a negative definite metric. After our discussion in the comments, I think I have finally cleaned up the relevant signs.
Fix a root $\alpha$ of $T$ in $\mathfrak g_{\mathbb C}$. Put $i_\alpha = -\lambda\lvert\lambda\rvert^{-1}$, where $\lambda = r(\mathrm d\alpha^\vee(1))$ ($H_\alpha \mathrel{:=} \mathrm d\alpha^\vee(1)$ is sometimes called the coroot, but I prefer to reserve that terminology for $\alpha^\vee$ itself), so that $i_\alpha$ is a square root of $-1$. Then $J$ carries $\xi_{\mathcal O_r}(r)$, where $\xi = X_\alpha + \overline{X_\alpha}$, to $\xi'_{\mathcal O_r}(r)$, where $\xi' = i_\alpha(X_\alpha - \overline{X_\alpha})$, for every $X_\alpha \in (\mathfrak g_{\mathbb C})_\alpha$.
$\DeclareMathOperator{\Tr}{Tr}
\DeclareMathOperator{\ad}{ad}
\DeclareMathOperator{\Lie}{Lie}
\newcommand{\g}{{\mathfrak g}}
\newcommand{\z}{{\mathfrak z}}
\newcommand{\s}{{\mathfrak s}}
\newcommand{\O}{{\mathcal O}}
\newcommand{\wh}{\widehat}
\newcommand{\wt}{\widetilde}$
Let $G$ be a connected compact Lie group, and let $\g$ denote its Lie algebra.
Write $\z$ for the center of $\g$ and set $\s=[\g,\g]$, which is a semisimple Lie algebra.
Then
$$ \g=\z\oplus\s.$$
Write $Z=Z(G)^0$ (the identity component of the center of $G$), $\ \wt S=[G,G]$, $S=G/Z(G)$.
Then $Z$ is a torus, whereas $\wt S$ and $S$ are semisimple Lie groups.
We can identify
$$ \z=\Lie Z,\quad \Lie S=\s=\Lie\wt S.$$
The group $G$ acts on $\g$ by adjoint representation.
Moreover, $G$ acts on $\g$ via the canonical surjective homomorphism
$\pi\colon G\to S$.
We write
$$ g\cdot X=s\cdot X,\quad\text{where} \ g\in G,\ s=\pi(g)\in S,\ X\in\g.$$
We write an element $X\in \g$ as
$$ X=X_\z+X_\s\ \quad \text{with}\ X_\z\in\z,\ X_\s\in \s\,.$$
Then
$$g\cdot X=X_\z+g\cdot X_\s=X_\z+s\cdot X_\s\,.$$
Let $\g^*$ denote the dual space for $\g$.
Then
$$\g^*=\z^*\oplus \s^*.$$
For $r\in \g^*$ we may write
$$r=r_\z+r_\s\quad\text{with}\ r_\z\in\z^*, \ r_\s\in \s^*.$$
Then for $g\in G$ we have
$$g\cdot r=r_\z+g\cdot r_\s=r_\z+s\cdot r_\s\,.$$
Let $X\in\g$, $\ X=X_\z+X_\s\,$. Then $[X_\z\,,X_\s]=0$.
It follows that
$$\exp -tX=(\exp -tX_\z)\cdot( \exp -tX_\s) \quad\text{with}\
\exp -tX_\z\in Z,\ \exp -tX_\s\in\wt S,$$
whence
$$(\exp -tX)\cdot \alpha=\alpha_\z+(\exp-t X_\s)\cdot \alpha_\s\quad
\text{for}\ \alpha=\alpha_\z+\alpha_\s\in \g^*.$$
Write
$$ \wh X=\frac d{dt}\Big|_{t=0}(\exp -tX)\cdot \alpha.$$
Then $\wh X=\wh{X_\s}$, where
$$ \wh {X_\s}=\frac d{dt}\Big|_{t=0}(\exp -tX_\s)\cdot \alpha_\s.$$
For $r=r_\z+r_\z \in\g^*$ write $\O_r=G\cdot r$, $\ \O_{r_\s}=S\cdot r_\s\,$. Then
$$ \O_r=r_\z+S\cdot r_\s=r_\z+\O_{r_\s}\,.$$
Thus for $\alpha=g\cdot r\in \O_r$ we have
\begin{equation}\label{e:*}
T_\alpha(\O_r)\cong T_{\alpha_\s}(\O_{r_\s}).\tag{$*$}
\end{equation}
Consider the adjoint representation
$$\ad\colon \g\to\mathfrak{gl}(\g),\quad (\ad X)\cdot Y=[X,Y].$$
For $X=X_\z+X_\s\in\g$ we have $\ad X=\ad X_\s\,$.
Consider the Killing form
$$ k\colon \g\times\g\to{\mathbb R}, \quad
(X,Y)\mapsto \Tr\!\big ((\ad X)\cdot (\ad Y)\big).$$
Then
$$ k(X,Y)=\Tr\!\big ((\ad X_\s)\cdot (\ad Y_\s)\big)=k(X_\s\,,Y_\s).$$
For $\lambda \in\g$, we define a skew-symmetric form on $T_\alpha(\O_r)$ by
$$\omega_\lambda(\wh X,\wh Y)=-k(\lambda,[X,Y]).$$
Since $[X,Y]=[X_\s\,,Y_\s]$, for $\lambda=\lambda_\z+\lambda_\s$ we have
$$\omega_\lambda(\wh X,\wh Y)=-k(\lambda,[X,Y])
=-k(\lambda_\s,[X_\s\,,Y_\s])=\omega_{\lambda_\s}(\wh{X_\s}\,, \wh{ Y_\s\, }\,).$$
We can identify the tangent spaces
$T_\alpha(\O_r)$ and $T_{\alpha_\s}(\O_{r_\s})$ by \eqref{e:*}.
Then the formula above is probably what you need.
Best Answer
This is true. One can use a few facts from Helgason's Differential Geometry, Lie Groups, and Symmetric Spaces to show that, indeed, $K = \mathrm{Stab}_G(Z)$.
Since $M=G/K$ is an irreducible Hermitian symmetric space, one can apply Theorem 6.1 from Chapter VIII of Helgason. The main point is that one can regard $G$ as a subgroup of $\mathrm{Aut}(\frak{g})$ with $K\subset G$ connected, and the circle $\mathrm{exp}(\mathbb{R}\cdot Z)\simeq S^1$ is the center of $K$. (See Helgason's Proposition 6.2.) Moreover, as Helgason explains, there will be a real number $t$ such that the element $z = \exp(tZ)$ is an element of order $2$ with the property that the $G$-centralizer of $z\in K$ is equal to $K$. Since the $G$-stabilizer of $Z$ is clearly contained in the $G$-centralizer of $z$, it follows that the $G$-stabilizer of $Z$ must equal $K$ as well.