Generalised binomial coefficients make sense when characteristic is not 2. $\newcommand{gbin}[1]{\binom{\frac{1}{2}}{#1}}\let\ge\geqslant$Consider the generalised binomial coefficient defined as
$$\gbin{n} :=  \frac{\left(\frac{1}{2}\right)\left(\frac{1}{2} - 1\right) \cdots \left(\frac{1}{2} - (n - 1)\right)}{n!}$$
for $n \ge 0$. (The value is $1$ when $n = 0$ by usual conventions.)

Is the following true: when $\gbin{n}$ is written in reduced form as $p/q$ with $q > 0$, then the only possible prime factor of $q$ is $2$.

(I have not picked up the question from some source, the reason why I think that the above is actually true is listed below.)

Motivation: If the above is true, then one would be able to define $\gbin{n}$ in any field $k$ whose characteristic is not $2$. Consequently, we would have the identity
$$\sqrt{1 + X} = \sum_{n \ge 0} \gbin{n} X^{n}$$
in the ring $k[\![X]\!]$ just like we have in $\Bbb Q[\![X]\!]$. Using Hensel lifts, we actually do have a square root of $1 + X$ in $k[\![X]\!]$. I believe the "canonical" square root obtained by starting with $1$ should be the one that we get when $k = \Bbb Q$.
Attempt:
Recursively, we have
$$\gbin{n + 1} = \frac{\frac{1}{2} - n}{n + 1}\gbin{n}.$$
However, it is not the case that the fraction $\frac{\frac{1}{2} - n}{n + 1}$ is always of the form $a/2^k$ and so this seems useless.
With some more effort, this answer shows that we have
$$\gbin{n} = \frac{(-1)^{n-1}}{2^{2n-1}n}\binom{2n-2}{n-1}.$$
This seems more promising as the question is then reduced to showing that
$$n \mid \binom{2n-2}{n-1}.$$

Yes, the denominators are always powers of $2$.
For a quick answer, you should see OEIS A046161. The $n$th entry of this sequence is the denominator in the $n$th term of the power series of $(1+x)^{k/2}$ for any odd $k$. In particular, when $k=1$ we get your series of interest.
Now, in the "formulas" section of this OEIS entry, it's shown that the $n$th entry of this sequence is
$2^{b_n}$, where $b_n$ is A005187. In particular, they're always powers of $2$.
In general, given any sequence of integers, you should just check if the OEIS has information that will solve your problem without you needing to think (as indeed it did here). If your sequence isn't in the OEIS, or if you find something new about your sequence, then you should add that information so that future mathematicians can have an easier job!

For a more direct answer, though, we can show that $n \mid \binom{2n-2}{n-1}$, which will give a proof based on what you've already noticed in your question. Indeed, it's well known that the Catalan Numbers (which are all integers) are given by the formula
$$
C_n = \frac{1}{n+1} \binom{2n}{n}
$$
so that $C_{n-1} = \frac{1}{n} \binom{2n-2}{n-1}$ is always an integer, as desired.

I hope this helps ^_^
