I think I found an answer, but I would appreciate if someone would check to see if I did this correctly before I mark this question as answered.
Under Series formulas for the Incomplete Beta Function, Wolfram gives that
$$B(z; a, b) \propto \frac{z^a}{a}\left(1+O(z)\right)$$
and under Approximation for the Beta Function Wikipedia gives that
$$B(a, b) \propto \Gamma(b)a^{-b}$$
when $a$ grows and $b$ stays fixed (credit to Brevan Ellefsen for finding this second approximation). Because of these two approximations we can write our limit as
$$\lim_{n\to\infty} nx^{n-1}\frac{C\frac{\left(1-\frac{x^2}{4}\right)^{\frac{n+1}{2}}}{\frac{n+1}{2}}}{\Gamma\left(\frac{1}{2}\right)\left(\frac{n+1}{2}\right)^{-\frac{1}{2}}}$$
where $C$ is some constant not depending on $n$.
$$\lim_{n\to\infty} nx^{n-1}\frac{C\left(1-\frac{x^2}{4}\right)^{\frac{n+1}{2}}}{\sqrt{\pi}\sqrt{\frac{n+1}{2}}}$$
$$c\lim_{n\to\infty} \frac{n}{\sqrt{n+1}}x^{n-1}\left(1-\frac{x^2}{4}\right)^{\frac{n+1}{2}}$$
for some other constant $c$.
Let $x = 2\sin\theta$. The limit then becomes
$$c\lim_{n\to\infty} \frac{n}{\sqrt{n+1}}\left(2\sin\theta\right)^{n-1}\left(\cos\theta\right)^{n+1}$$
$$c\lim_{n\to\infty} \frac{n}{\sqrt{n+1}}\left(2\sin\theta\cos\theta\right)^{n-1}\cos^2\theta$$
$$c\lim_{n\to\infty} \frac{n}{\sqrt{n+1}}\left(\sin\left(2\theta\right)\right)^{n-1}\cos^2\theta$$
Note that, whenever $\left|\sin\left(2\theta\right)\right| < 1$, the $\left(\sin\left(2\theta\right)\right)^{n-1}$ term tends to $0$ more quickly than the $\frac{n}{\sqrt{n+1}}$ term tends to $\infty$ (since $\frac{n}{\sqrt{n+1}}$ is asymptotically a power of $n$ and $\left(\sin\left(2\theta\right)\right)^{n-1}$ is exponential), and the limit is just $0$; also note that when $\left|\sin\left(2\theta\right)\right| = 1$ the limit tends to $\infty$. Thus, this reduces to the problem of finding $\theta$ such that
$$\sin^2\left(2\theta\right) = 1$$
$$4\sin^2\theta\cos^2\theta = 1$$
$$x^2\left(1-\frac{x^2}{4}\right) = 1$$
$$4x^2-x^4 = 4$$
$$x^2 = 2$$
$$x = \pm\sqrt{2}$$
So, at these points the limit tends to $\infty$ whereas everywhere else it tends to $0$.
Using the hypergeometric representation of the incomplete Beta function
\begin{equation}
B_x\left( c,d \right)=\frac{x^c}{c}{}_2F_1\left( c,1-d;1+c;x \right)
\end{equation}
the integral can be written as
\begin{align}
I\left( a,b,c,d \right)&=\int_{0}^1x^{a-1}(1-x)^{b-1}B_x(c,d)\,dx\\
&=\frac{1}{c}\int_{0}^1x^{a+c-1}(1-x)^{b-1}{}_2F_1\left( c,1-d;1+c;x \right)\,dx\\
&=\frac{1}{c}\sum_{k=0}^\infty \frac{(c)_k(1-d)_k}{(1+c)_kk!} \int_0^1x^{a+c+k-1}(1-x)^{b-1}\,dx\\
&=\frac{1}{c}\sum_{k=0}^\infty \frac{(c)_k(1-d)_k}{(1+c)_kk!} \frac{\Gamma(b)\Gamma(a+c+k)}{\Gamma(a+b+c+k)}\\
&=\frac{1}{c}\frac{\Gamma(b)\Gamma(a+c)}{\Gamma(a+b+c)}\sum_{k=0}^\infty \frac{(c)_k(1-d)_k}{(1+c)_kk!} \frac{(a+c)_k}{(a+b+c)_k}\\
&=\frac{1}{c}\frac{\Gamma(b)\Gamma(a+c)}{\Gamma(a+b+c)}\,{}_3F_2\left( 1-d,a+c,c;1+c,a+b+c ;1\right)
\end{align}
Using this identity for the generalized hypergeometric function:
\begin{equation}
{}_3F_2\left( a_1,a_2,a_3;b_1,b_2;1 \right)=\frac{\Gamma(b_1)\Gamma(b_1+b_2-a_1-a_2-a_3)}{\Gamma(b_1-a_1)\Gamma(b_1+b_2-a_2-a_3)}{}_3F_2\left( a_1,b_2-a_2,b_2-a_3;b_2,b_1+b_2-a_2-a_3;1 \right)
\end{equation}
Here we choose $a_1=1-d,a_2=a+c,a_3=c,b_1=a+b+c,b_2=1+c$ to obtain
\begin{align}
I\left( a,b,c,d \right)&=\frac{1}{c}\frac{\Gamma(b)\Gamma(a+c)}{\Gamma(a+b+c)}\frac{\Gamma(a+b+c)\Gamma(b+d)}{\Gamma(a+b+c+d-1)\Gamma(b+1)}\,{}_3F_2\left( 1-d,1-a,1;1+c,1+b ;1\right)\\
&=\frac{1}{bc}\frac{\Gamma(a+c)\Gamma(b+d)}{\Gamma(a+b+c+d-1)}\,{}_3F_2\left( 1,1-a,1-d;1+b,1+c ;1\right)\\
&=\frac{a+b+c+d-1}{bc}B(a+c,b+d)\,{}_3F_2\left( 1,1-a,1-d;1+b,1+c ;1\right)
\end{align}
which gives simple results if $a$ or $d$ are positive integer, as the hypergeometric series is finite: as $1-a\le0$ or $1-b\le0$, in the series definition of the hypergeometric function, there are $\operatorname{min}(a,b)$ terms (numerator of the coefficients cancel after that). One can also check the given result when $a=c$ and $b=d$.
Best Answer
For $x\to 0+$, we can expand the factor $(1-t)^{\beta-1}$ about $t=0$ to find $$ I_x (\alpha ,\beta ) = \frac{{\Gamma (\alpha + \beta )}}{{\Gamma (\alpha )\Gamma (\beta )}}\frac{{x^\alpha }}{\alpha }\left( {1 + \mathcal{O}(x)} \right). $$ With your specific parameters, this gives $$ \frac{b}{{a + b}}\left( {\frac{{\log ^q n}}{{\sqrt n }}} \right)^{a/\log n} \left( {1 + \mathcal{O}\!\left( {\frac{{\log ^q n}}{{\sqrt n }}} \right)} \right) = \frac{{b\, \mathrm{e}^{ - a/2} }}{{a + b}}\left( {1 + \mathcal{O}\!\left( {\frac{{\log \log n}}{{\log n}}} \right)} \right), $$ as $n\to +\infty$.