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$\newcommand\C{\mathbb C}\newcommand\R{\mathbb R}$Yes, this follows by Loewner's theorem on monotone matrix functions (see e.g. Theorem 1.6), which in particular implies the following:

Let $M_n$ denote the set of all analytic functions $f\colon\C\setminus(-\infty,0]\to\C$ such that $f((0,\infty))\subseteq\R$ and $$A\le B\implies f(A)\le f(B)$$ for all $n\times n$ positive-definite matrices $A$ and $B$, where $A\le B$ means that $B-A$ is positive semidefinite. Then $f\in M_n$ for all natural $n$ if $$\Im z>0\implies \Im f(z)>0.$$

The above conditions on $f$ hold if $f(z)=z^\alpha$ for $\alpha\in(0,1]$ and all $z\in\C\setminus(-\infty,0]$.

So, your desired result immediately follows. (Even more immediately, your desired result follows from Theorem 4.1.)

$\newcommand\C{\mathbb C}\newcommand\R{\mathbb R}\newcommand\al{\alpha}$Yes, this follows by Loewner's theorem on monotone matrix functions (see e.g. Theorem 1.6), which in particular implies the following:

Let $M_n$ denote the set of all analytic functions $f\colon\C\setminus(-\infty,0]\to\C$ such that $f((0,\infty))\subseteq\R$ and $$A\le B\implies f(A)\le f(B)$$ for all $n\times n$ positive-definite matrices $A$ and $B$, where $A\le B$ means that $B-A$ is positive semidefinite. Then $f\in M_n$ for all natural $n$ if $$\Im z>0\implies \Im f(z)>0.$$

The above conditions on $f$ hold if $f(z)=z^\alpha$ for $\alpha\in(0,1]$ and all $z\in\C\setminus(-\infty,0]$.

So, your desired result immediately follows.

Even more immediately, your desired result follows from Theorem 4.1, which in turn follows from the identity $$x^\al=\frac{\sin\pi\al}\pi\int_0^\infty w^{\al-1}x(x+w)^{-1}\,dw$$ for real $x>0$, since in this identity $x$ can be replaced by any positive-definite matrix $A$ (see formula (4.5)), and the monotonicity of $A(A+w)^{-1}=I-w(A+w)^{-1}$ in $A$ for $w>0$ is easy to check (say, by differentiation).

$\newcommand\C{\mathbb C}\newcommand\R{\mathbb R}$Yes, this follows by Loewner's theorem on monotone matrix functions (see e.g. Theorem 1.6), which in particular implies the following:

Let $M_n$ denote the set of all analytic functions $f\colon\C\setminus(-\infty,0]\to\C$ such that $f((0,\infty))\subseteq\R$ and $$A\le B\implies f(A)\le f(B)$$ for all $n\times n$ positive-definite matrices $A$ and $B$, where $A\le B$ means that $B-A$ is positive semidefinite. Then $f\in M_n$ for all natural $n$ iff $$\Im z>0\implies \Im f(z)>0.$$

The above conditions on $f$ hold if $f(z)=z^\alpha$ for $\alpha\in(0,1]$ and all $z\in\C\setminus(-\infty,0]$.

So, your desired result immediately follows.

$\newcommand\C{\mathbb C}\newcommand\R{\mathbb R}$Yes, this follows by Loewner's theorem on monotone matrix functions (see e.g. Theorem 1.6), which in particular implies the following:

Let $M_n$ denote the set of all analytic functions $f\colon\C\setminus(-\infty,0]\to\C$ such that $f((0,\infty))\subseteq\R$ and $$A\le B\implies f(A)\le f(B)$$ for all $n\times n$ positive-definite matrices $A$ and $B$, where $A\le B$ means that $B-A$ is positive semidefinite. Then $f\in M_n$ for all natural $n$ if $$\Im z>0\implies \Im f(z)>0.$$

The above conditions on $f$ hold if $f(z)=z^\alpha$ for $\alpha\in(0,1]$ and all $z\in\C\setminus(-\infty,0]$.

So, your desired result immediately follows. (Even more immediately, your desired result follows from Theorem 4.1.)

$\newcommand\C{\mathbb C}\newcommand\R{\mathbb R}$Yes, this follows by Loewner's theorem on monotone matrix functions (see e.g. Theorem 1.6), which in particular implies the following:

Let $M_n$ denote the set of all analytic functions $f\colon\C\setminus(-\infty,0]\to\C$ such that $f((0,\infty))\subseteq\R$ and $$A\le B\implies f(A)\le f(B)$$ for all $n\times n$ positive-definite matrices $A$ and $B$, where $A\le B$ means that $B-A$ is positive semidefinite. Then $f\in M_n$ for all natural $n$ iff $$\Im z>0\implies \Im f(z)>0.$$

The above conditions on $f$ hold if $f(z)=z^\alpha$ for $\alpha\in[0,1]$ and all $z\in\C\setminus(-\infty,0]$.

So, your desired result immediately follows.

$\newcommand\C{\mathbb C}\newcommand\R{\mathbb R}$Yes, this follows by Loewner's theorem on monotone matrix functions (see e.g. Theorem 1.6), which in particular implies the following:

Let $M_n$ denote the set of all analytic functions $f\colon\C\setminus(-\infty,0]\to\C$ such that $f((0,\infty))\subseteq\R$ and $$A\le B\implies f(A)\le f(B)$$ for all $n\times n$ positive-definite matrices $A$ and $B$, where $A\le B$ means that $B-A$ is positive semidefinite. Then $f\in M_n$ for all natural $n$ iff $$\Im z>0\implies \Im f(z)>0.$$

The above conditions on $f$ hold if $f(z)=z^\alpha$ for $\alpha\in(0,1]$ and all $z\in\C\setminus(-\infty,0]$.

So, your desired result immediately follows.