A uniform asymptotic expansion can be obtained using the result linked in the comments. However, I will now derive simpler asymptotic approximations. I will set $A = a$, $x^2 = x$, and assume $a > 0$ and $0 < x \leq 1$. Substituting $u = 1 - \mathrm{e}^{-t}$ into the integral formula provided in Svyatoslav's answer gives
$$
M(a, n + 1, nx) = \frac{\Gamma(n+1)}{\Gamma(n+1-a)} \frac{1}{\Gamma(a)} \int_0^{+\infty} (\mathrm{e}^t - 1)^{a-1} \mathrm{e}^{-n(t - x(1 - \mathrm{e}^{-t}))} \, \mathrm{d}t.
$$
Case 1. For $0 \le x < 1$ and $(1 - x)^2 n \to +\infty$, Laplace's method yields the asymptotic expansion:
$$
\frac{1}{\Gamma(a)} \int_0^{+\infty} (\mathrm{e}^t - 1)^{a-1} \mathrm{e}^{-n(t - x(1 - \mathrm{e}^{-t}))} \, \mathrm{d}t \sim \frac{1}{(1 - x)^a} \frac{1}{n^a} \sum_{k=0}^\infty \frac{c_k(x;a)}{n^k}
$$
as $n \to +\infty$, where the coefficients $c_k(x;a)$ are given by
$$
c_k(x;a) = \frac{1}{(1 - x)^k} \frac{(a)_k}{k!} \left[ \frac{\mathrm{d}^k}{\mathrm{d}t^k} \left( \left( \frac{\mathrm{e}^t - 1}{t} \right)^{a-1} \left( \frac{(1 - x)t}{t - x(1 - \mathrm{e}^{-t})} \right)^{a+k} \right) \right]_{t=0}.
$$
In particular,
$$
c_0(x;a) = 1, \quad c_1(x;a) = - \frac{a(2ax - a + 1)}{2(1 - x)^2}.
$$
Using the known asymptotic expansion for the ratio of two gamma functions, we find
$$
\frac{\Gamma(n+1)}{\Gamma(n+1-a)} \sim n^a \sum_{k=0}^\infty \binom{a-1}{k} B_k^{(a)} \frac{1}{n^k}
$$
as $n \to +\infty$, where $B_k^{(a)}$ are the generalised Bernoulli numbers. Multiplying these asymptotic expansions gives
$$ \tag{1}
M(a, n+1, nx) \sim \frac{1}{(1-x)^a} \sum_{k=0}^\infty \frac{d_k(x;a)}{n^k}
$$
as $n \to +\infty$, where
$$
d_k(x;a) = \sum_{j=0}^k \binom{a-1}{j} B_j^{(a)} c_{k-j}(x;a).
$$
In particular,
$$
d_0(x;a) = 1, \quad d_1(x;a) = -\frac{ax((a-1)x + 2)}{2(1 - x)^2}.
$$
The coefficients $d_k(x;a)$ have poles of order $2k$ at $x = 1$, which implies that we require $(1-x)^2 n \to +\infty$.
Case 2. For $x \approx 1$, we examine $M(a, n + 1, n + \tau n^{1/2})$:
\begin{align*}
M(a,n + 1,n + \tau n^{1/2} ) =\; &\frac{{\Gamma (n + 1)}}{{\Gamma (n + 1 - a)}}\frac{1}{{\Gamma (a)}} \\ &\times \int_0^{ + \infty } {({\rm e}^t - 1)^{a - 1} \exp ( - n({\rm e}^{ - t}+t - 1) + n^{1/2} \tau (1 - {\rm e}^{ - t} ))\,{\rm d}t}.
\end{align*}
Here, $\tau$ is a fixed real number. Proceeding similarly to $\S4.1$ of this paper, an extension of Laplace's method yields
\begin{align*} \tag{2}
M(a,n + 1,n + \tau n^{1/2} ) \sim \; & n^{a/2}\exp \left( {\frac{{\tau ^2 }}{4}} \right) U\!\left( {a - \frac{1}{2}, - \tau } \right)\sum\limits_{k = 0}^\infty {\frac{{P_k (\tau ;a)}}{{n^{k/2} }}} \\ & + n^{(a - 1)/2} \exp \left( {\frac{{\tau ^2 }}{4}} \right)U\!\left( {a + \frac{1}{2}, - \tau } \right)\sum\limits_{k = 0}^\infty {\frac{{Q_k (\tau ;a)}}{{n^{k/2} }}}
\end{align*}
as $n \to +\infty$. Here, the $U$ denotes the parabolic cylinder function. The coefficients $P_k(\tau; a)$ and $Q_k(\tau; a)$ are polynomials in $\tau$ and $a$, with $P_0(\tau; a)=1$. By inspection, they are both $\mathcal{O}(\tau^{3k+2})$ for large $\tau$, which allows us to assume $\tau = o(n^{1/6})$. With this assumption, the two cases together cover $0 \le x \leq 1$.
Update. The leading order of the asymptotics in $(2)$ can be re-expressed as
$$\tag{3}
M(a, n + 1, nx) \sim n^{a/2} \exp\left( \frac{{(1 - x)^2}}{4}n \right) U\!\left( a - \frac{1}{2}, n^{1/2} (1 - x) \right)
$$
as $n \to +\infty$, provided $x = 1 + o(n^{-1/3})$. However, when $x = 1 + \mathcal{O}(n^{-1/2 + \varepsilon})$, the right-hand side asymptotically behaves like $(1 - x)^{-a}$, aligning with $(1)$. Therefore, $(3)$ holds for $0 \leq x \leq 1$.
A numerical example with $a = 2$, $n = 10^4$, and $0 \leq x \leq 1$ is shown in the figure below. The red curve represents the exact values, while the dashed blue curve corresponds to the approximation in $(3)$.
