Your second guess is also correct. 

At first, we write down the sign of a permutation as $\prod_{1\leqslant i<j\leqslant p-1}\frac{a^j-a^i}{j-i}$ modulo $p$. The denominator equals $(p-2)!(p-3)!\dots 1!$, and denoting $p=2m+1$ ($m$ is odd) we write it as $m!\prod_{j=1}^{m-1} j!(p-1-j)!=m!\prod_{j=1}^{m-1} (-1)^{j+1}=m!(-1)^{(m-1)/2}$. Here $m!$ appears.

Now for the numerator. We consider the polynomial $F(x)=\prod_{1\leqslant i<j\leqslant p-1} (x^j-x^i)$ modulo the cyclotomic polynomial $\Phi_{p-1}(x)$. We may write $F(x)=h(x)\Phi_{p-1}(x)+r(x)$ for polynomials $h,r$ with integer coefficients, $\deg r<\varphi(p-1)$, and of course $F(a)$ and $r(a)$ are congruent modulo $p$. So our goal is to find $r$. For this we substitute the *complex* root $\omega=e^{2\pi i/(p-1)}$ of $\Phi_{p-1}$ and get $r(\omega)=F(\omega)$. $F(\omega)$ may be calculated by the following (not original, used for calculating the sign of Gauss sums) trick: at first, we know the argument of $F(\omega)$, since we know the argument of each bracket $\omega^k-\omega^j$ and they are easily summed up. At second, we know $F^2(\omega)$. Indeed, using the formula 
$$\prod_{j\ne i} (\omega^i-\omega^j)={\frac{t^{p-1}-1}{t-\omega^i}}|_{t=\omega^i}=-(p-1)\omega^{-i}$$
we get

$$
F^2(\omega)=-\prod_i\prod_{j\ne i} (\omega^i-\omega^j)=-(p-1)^{p-1}\omega^{-p(p-1)/2}=(p-1)^{p-1}.
$$

Thus $r(\omega)=F(\omega)=\pm (p-1)^{(p-1)/2}$, where the sign may be obtained from the argument of $F(\omega)$. Since $\deg r<\varphi(p-1)=\deg \Phi_{p-1}$ and $\Phi_{p-1}$ is irreducible, we conclude that $r=\pm (p-1)^{(p-1)/2}$ identically and modulo $p$ we get $r(a)=\mp$. Probably this sign depends on the parity of $(m-1)/2=(p-3)/4$ (that is easy to check in any case) and totally thу sign of $\sigma_a$ is $m!$ modulo $p$.