Denote the numerator by $I(a)$. The change of variables $x_i=ay_i$ gives
$$
I(a)=a^{n(\alpha-\frac{n+1}2)+\frac{n(n+1)}2}\times \\
\times\int_{[0,1]^n}\prod_{i=1}^ny_i^{\alpha-\frac{n+1}{2}}\prod_{i=1}^n(1-ay_i)^{-1/2}\prod_{i<j}|y_i-y_j|.
$$
The latter integral may be estimated using the two-sided estimates $1\leqslant (1-ay_i)^{-1/2}\leqslant (1-y_i)^{-1/2}$. So, we get $$c_1(n) a^{n(\alpha-\frac{n+1}2)+\frac{n(n+1)}2}\leqslant I(a)\leqslant c_2(n) a^{n(\alpha-\frac{n+1}2)+\frac{n(n+1)}2},$$
where $c_1(n)$ and $c_2(n)$ are corresponding integrals (actually $c_2(n)=I(1)$).
It implies
$$
\frac{I(a)}{I(b)}\leqslant \frac{c_2(n)}{c_1(n)}\cdot (a/b)^{n(\alpha-\frac{n+1}2)+\frac{n(n+1)}2}.
$$
You may get some explicit bounds for $c_2(n)/c_1(n)$
, for example as follows.
We have
$$
c_2(n)=I(1)=\int_{[0,1]^n}\prod_{i=1}^ny_i^{\alpha-\frac{n+1}{2}}\prod_{i=1}^n(1-y_i)^{-1/2}\prod_{i<j}|y_i-y_j| dy_1\ldots dy_n,
$$
and $c_1(n)$ the same integral without $\prod_{i=1}^n(1-y_i)^{-1/2}$. In the integral for $c_2(n)$, change the variables: take $y_i=1-t_i^2$. You get
$$
c_2(n)=2^n\int_{[0,1]^n}\prod_{i=1}^n(1-t_i^2)^{\alpha-\frac{n+1}{2}}\prod_{i<j}|t_i^2-t_j^2| dt_1\ldots dt_n.
$$
Use the bounds $1-t_i^2=(1-t_i)(1+t_i)\leqslant 2(1-t_i)$, $|t_i^2-t_j^2|=|t_i-t_j|(t_i+t_j)\leqslant 2|t_i-t_j|$. We get
$$
c_2(n)\leqslant 2^{n(\alpha-\frac{n+1}{2})+\frac{n(n+1)}2}
\int_{[0,1]^n}\prod_{i=1}^n(1-t_i)^{\alpha-\frac{n+1}{2}}\prod_{i<j}|t_i-t_j| dt_1\ldots dt_n=\\
=2^{n(\alpha-\frac{n+1}{2})+\frac{n(n+1)}2} c_1(n),
$$
the last equality follows from the change of variables $1-t_i=z_i$.
Is it ok for you or you need the estimates which are explicit w.r.t. $n$?
