Only the symmetric part of $A$ contributes to the integrand, so we may assume $A$ is symmetric and diagonalize it as $A=O\Lambda O^T$ with $O$ orthogonal and $\Lambda={\rm diag}\,(\lambda_1,\lambda_2,\ldots\lambda_N)$. The orthogonal matrix may be removed by a change of variables, so without loss of generality we can take a diagonal $A$. So we seek the integral $$I=\int \prod_{i, j=1}^{N} d J_{i j} \exp \left\{-\frac{N}{2(1-\tau^2)}\left( \sum_{i, j} \lambda_j J_{ij}^2-\tau\sum_{ij}J_{ij}J_{ji}\right)\right\}.$$ I now proceed similarly to here. Decompose the sum over $i,j$ as $$\sum_{ij}\lambda_j J_{ij}^2-\tau\sum_{ij}J_{ij}J_{ji}=$$ $$\qquad\qquad=\sum_{i}(\lambda_i-\tau)J_{ii}^2+\sum_{i<j}\left(\lambda_j J_{ij}^2+\lambda_i J_{ji}^2-2\tau J_{ij}J_{ji}\right)$$ $$\qquad\qquad\equiv\sum_{i} A_i+\sum_{i<j} B_{ij}.$$ Then perform the Gaussian integrals separately for each term in the sum, $$I=\left(\prod_{i=1}^N\int e^{-\beta A_i}dJ_{ii}\right)\left(\prod_{i<j=1}^{N}\int\int e^{-\beta B_{ij}}dJ_{ij}dJ_{ji}\right)$$ $$\qquad\qquad=(\pi/\beta)^{N^2/2}\left(\prod_{i=1}^N(\lambda_i-\tau)^{-1/2}\right)\left(\prod_{i<j=1}^N(\lambda_i\lambda_j-\tau^2)^{-1/2}\right),$$ where I have defined $\beta=\frac{1}{2}N(1-\tau^2)^{-1}$, and assumed that $\beta>0$, $\lambda_i>\tau$ for all $i$.

So knowledge of only the determinant of $A$ is not sufficient to evaluate the integral, you need to know the individual eigenvalues -- not just their product.

Only the symmetric part of $A$ contributes to the integrand, so we may assume $A$ is symmetric and diagonalize it as $A=O\Lambda O^T$ with $O$ orthogonal and $\Lambda={\rm diag}\,(\lambda_1,\lambda_2,\ldots\lambda_N)$. The orthogonal matrix may be removed by a change of variables, so without loss of generality we can take a diagonal $A$. So we seek the integral $$I=\int \prod_{i, j=1}^{N} d J_{i j} \exp \left\{-\frac{N}{2(1-\tau^2)}\left( \sum_{i, j} \lambda_j J_{ij}^2-\tau\sum_{ij}J_{ij}J_{ji}\right)\right\}.$$ I now proceed similarly to here. Decompose the sum over $i,j$ as $$\sum_{ij}\lambda_j J_{ij}^2-\tau\sum_{ij}J_{ij}J_{ji}=$$ $$\qquad\qquad=\sum_{i}(\lambda_i-\tau)J_{ii}^2+\sum_{i<j}\left(\lambda_j J_{ij}^2+\lambda_i J_{ji}^2-2\tau J_{ij}J_{ji}\right)$$ $$\qquad\qquad\equiv\sum_{i} A_i+\sum_{i<j} B_{ij}.$$ Then perform the Gaussian integrals separately for each term in the sum, $$I=\left(\prod_{i=1}^N\int e^{-\beta A_i}dJ_{ii}\right)\left(\prod_{i<j=1}^{N}\int\int e^{-\beta B_{ij}}dJ_{ij}dJ_{ji}\right)$$ $$\qquad\qquad=(\pi/\beta)^{N^2/2}\left(\prod_{i=1}^N(\lambda_i-\tau)^{-1/2}\right)\left(\prod_{i<j=1}^N(\lambda_i\lambda_j-\tau^2)^{-1/2}\right),$$ where I have defined $\beta=\frac{1}{2}N(1-\tau^2)^{-1}$, and assumed that $\beta>0$, $\lambda_i>\tau$ for all $i$.

So knowledge of only the determinant of $A$ is not sufficient to evaluate the integral, you need to know the individual eigenvalues -- not just their product. In particular, if $\tau$ happens to be close to one particular eigenvalue of $A$, then that eigenvalue will dominate the integral.

Only the symmetric part of $A$ contributes to the integrand, so we may assume $A$ is symmetric and diagonalize it as $A=O\Lambda O^T$ with $O$ orthogonal and $\Lambda={\rm diag}\,(\lambda_1,\lambda_2,\ldots\lambda_N)$. The orthogonal matrix may be removed by a change of variables, so without loss of generality we can take a diagonal $A$. So we seek the integral $$I=\int \prod_{i, j=1}^{N} d J_{i j} \exp \left\{-\frac{N}{2(1-\tau^2)}\left( \sum_{i, j} \lambda_j J_{ij}^2-\tau\sum_{ij}J_{ij}J_{ji}\right)\right\}.$$ Decompose the sum over $i,j$ as $$\sum_{ij}\lambda_j J_{ij}^2-\tau\sum_{ij}J_{ij}J_{ji}=$$ $$\qquad\qquad=\sum_{i}(\lambda_i-\tau)J_{ii}^2+\sum_{i<j}\left(\lambda_j J_{ij}^2+\lambda_i J_{ji}^2-2\tau J_{ij}J_{ji}\right)$$ $$\qquad\qquad\equiv\sum_{i} A_i+\sum_{i<j} B_{ij}.$$ Then perform the Gaussian integrals separately for each term in the sum, $$I=\left(\prod_{i=1}^N\int e^{-\beta A_i}dJ_{ii}\right)\left(\prod_{i<j=1}^{N}\int\int e^{-\beta B_{ij}}dJ_{ij}dJ_{ji}\right)$$ $$\qquad\qquad=(\pi/\beta)^{N^2/2}\left(\prod_{i=1}^N\frac{1 }{\lambda_i-\tau}\right)^{1/2}\left(\prod_{i<j=1}^N\frac{1 }{\lambda_i\lambda_j-\tau^2}\right)^{1/2},$$ where I have defined $\beta=\frac{1}{2}N(1-\tau^2)^{-1}$, and assumed that $\beta>0$, $\lambda_i>\tau$ for all $i$.

Only the symmetric part of $A$ contributes to the integrand, so we may assume $A$ is symmetric and diagonalize it as $A=O\Lambda O^T$ with $O$ orthogonal and $\Lambda={\rm diag}\,(\lambda_1,\lambda_2,\ldots\lambda_N)$. The orthogonal matrix may be removed by a change of variables, so without loss of generality we can take a diagonal $A$. So we seek the integral $$I=\int \prod_{i, j=1}^{N} d J_{i j} \exp \left\{-\frac{N}{2(1-\tau^2)}\left( \sum_{i, j} \lambda_j J_{ij}^2-\tau\sum_{ij}J_{ij}J_{ji}\right)\right\}.$$ I now proceed similarly to here. Decompose the sum over $i,j$ as $$\sum_{ij}\lambda_j J_{ij}^2-\tau\sum_{ij}J_{ij}J_{ji}=$$ $$\qquad\qquad=\sum_{i}(\lambda_i-\tau)J_{ii}^2+\sum_{i<j}\left(\lambda_j J_{ij}^2+\lambda_i J_{ji}^2-2\tau J_{ij}J_{ji}\right)$$ $$\qquad\qquad\equiv\sum_{i} A_i+\sum_{i<j} B_{ij}.$$ Then perform the Gaussian integrals separately for each term in the sum, $$I=\left(\prod_{i=1}^N\int e^{-\beta A_i}dJ_{ii}\right)\left(\prod_{i<j=1}^{N}\int\int e^{-\beta B_{ij}}dJ_{ij}dJ_{ji}\right)$$ $$\qquad\qquad=(\pi/\beta)^{N^2/2}\left(\prod_{i=1}^N(\lambda_i-\tau)^{-1/2}\right)\left(\prod_{i<j=1}^N(\lambda_i\lambda_j-\tau^2)^{-1/2}\right),$$ where I have defined $\beta=\frac{1}{2}N(1-\tau^2)^{-1}$, and assumed that $\beta>0$, $\lambda_i>\tau$ for all $i$.

So knowledge of only the determinant of $A$ is not sufficient to evaluate the integral, you need to know the individual eigenvalues -- not just their product.