Interesting question. We may start from the definition of the Beta function:
$$B(m, n) = \int_0^1 t^{m-1}(1-t)^{n-1}\ \text{d}t$$
and rewrite it when $m\to 2m$:
$$B(2m, n) = \int_0^1 t^{2m-1}(1-t)^{n-1}\ \text{d}t$$
$$B(2m, n) = \int_0^1 t^{2m-1} \frac{(1-t)^n}{1-t}\ \text{d}t$$
Now, since the range of integration is $[0, 1]$, we are allowed to make use of the geometric series
$$\frac{1}{1-t} = \sum_{k = 0}^{+\infty} t^k$$
Hence
$$B(2m, n) = \int_0^1 t^{2m-1} (1-t)^n \sum_{k = 0}^{+\infty} t^k\ \text{d}t = \sum_{k = 0}^{+\infty} \int_0^1 t^{2m-1} (1-t)^n t^k\ \text{d}t$$
And easily write:
$$\sum_{k = 0}^{+\infty} \int_0^1 t^{2m-1+k} (1-t)^n\ \text{d}t$$
Calling now
$$2m-1+k = a ~~~~~~~~~~~ n = b$$
We notice that the integral is well known:
$$ \int_0^1 t^a (1-t)^b\ \text{d}t \equiv B(a+1, b+1)$$
Then we end up with the partial result (re-expanding $a$ and $b$):
$$B(2m, n) = \sum_{k = 0}^{+\infty} B(2m+k, n+1)$$
That series does exist and it does converge to a known result:
$$\sum_{k = 0}^{+\infty} B(2m+k, n+1) = \frac{\Gamma (n) \Gamma (2 m+n+1) B(2 m,n+1)}{\Gamma (n+1) \Gamma (2 m+n)}$$
What you end up with is a sort of recursive relation for the Beta function:
$$B(2m, n) = \frac{\Gamma (n) \Gamma (2 m+n+1) B(2 m,n+1)}{\Gamma (n+1) \Gamma (2 m+n)}$$
BUT
The above expression can be simplified!
$$\frac{\Gamma (n) \Gamma (2 m+n+1) B(2 m,n+1)}{\Gamma (n+1) \Gamma (2 m+n)} \equiv \frac{\Gamma (2 m) \Gamma (n)}{\Gamma (2 m+n)}$$
What we obtained is actually nothing than what we would have obtained by simply substituting at the beginning $m\to 2m$ in the Gamma function / Beta function definition.
$$B(2m, n) = \frac{\Gamma (2 m) \Gamma (n)}{\Gamma (2 m+n)}$$
This really suggest that such a particular duplication formula for the Beta function may not exist at all, since all you need is the Gamma function and ITS duplication formula, through which you can evaluate $\Gamma(2m)$.
Seems like that this is the only "duplication formula" for the beta function.
(Also, I found nothing on reviews or literature).
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\begin{align}
&\bbox[10px,#ffd]{\lim _{x \to \infty}
\ln\pars{\root{x}\Gamma\pars{x/2} \over
\Gamma\pars{\bracks{x + 1}/2}}} =
{1 \over 2}\,\ln\pars{2} + \lim _{x \to \infty}
\ln\pars{\root{x}\Gamma\pars{x} \over \Gamma\pars{x + 1/2}}
\\[5mm] = &\
{1 \over 2}\,\ln\pars{2} + \lim _{x \to \infty}
\ln\pars{\root{x}\bracks{x - 1}! \over \bracks{x - 1/2}!}
\\[5mm] = &\
{1 \over 2}\,\ln\pars{2} + \lim _{x \to \infty}
\ln\pars{\root{x}\,{\root{2\pi}\bracks{x - 1}^{x - 1/2}
\expo{-\pars{x - 1}} \over
\root{2\pi}\bracks{x - 1/2}^{x}\expo{-\pars{x - 1/2}}}}
\\[5mm] = &\
{1 \over 2}\,\ln\pars{2} + \lim _{x \to \infty}
\ln\pars{\root{x}\,{x^{x - 1/2}\bracks{1 - 1/x}^{x - 1/2}
\root{\expo{}} \over
x^{x}\bracks{1 - \pars{1/2}/x}^{x}}}
\\[5mm] = &\
{1 \over 2}\,\ln\pars{2}\ +\
\underbrace{\ln\pars{{\expo{-1}\root{\expo{}} \over\expo{-1/2}}}}
_{\ds{=\ 0}}\ =\ \bbx{{1 \over 2}\,\ln\pars{2}}\approx 0.3466
\end{align}
Best Answer
When $q > 2$, see @Gary's comment.
When $1 < q \le 2$, we have \begin{align*} \frac{\Gamma(q)\Gamma(q + 1)}{\Gamma(2q + 1)} &= \int_0^1 t^q (1 - t)^{q - 1}\mathrm{d} t \\ &= \int_0^{1/2} t^q (1 - t)^{q - 1}\mathrm{d} t + \int_{1/2}^1 t^q (1 - t)^{q - 1}\mathrm{d} t \\ &= \int_0^{1/2} t^q (1 - t)^{q - 1}\mathrm{d} t + \int_0^{1/2} (1 - t)^q t^{q - 1}\mathrm{d} t \\ &= \int_0^{1/2} t^{q- 1} (1 - t)^{q - 1}\mathrm{d} t\\ &\le \int_0^{1/2} t^{q- 1} (1 - (q - 1)t)\mathrm{d} t\\ &= \frac{-q^2 + 3q + 2}{2^{q + 1}q (q + 1)} \end{align*} where we have used Bernoulli inequality to get $(1 - t)^{q - 1} \le 1 - (q - 1)t$. Thus, we have $$q 2^q \frac{\Gamma(q)\Gamma(q + 1)}{\Gamma(2q + 1)} \le \frac{-q^2 + 3q + 2}{2 (q + 1)} < 1.$$
We are done.