Suppose
$$ \Sigma^{-1} \sim \text{Wishart}_p \left(\frac{\Sigma_0^{-1}}{d_0},d_0\right) $$
That is $\Sigma$ has a inverse Wishart distribution.
Furthermore
$$ W \mid \Sigma \sim \operatorname{Wishart}_p(\Sigma, d) $$
$$ Z \mid \Sigma \sim \mathcal{N}_p(0,\Sigma) $$
and $W$ and $Z$ are conditionally independent given $\Sigma$.
I want to prove (or disprove) that $$ \frac{Z^T\Sigma^{-1}Z}{Z^T(W+\Sigma_0d_0)^{-1}Z} \sim \chi^2(d+d_0-p+1). $$
I can show that the result holds in a few special cases:
Special case 1: p=1 When $p=1$ the Wishart distributions reduces to $\chi^2$ distributions: $$\Sigma^{-1} \sim \frac{1}{d_0\Sigma_0}\chi^2(d_0)$$ $$ W \mid \Sigma \sim \Sigma\chi^2(d) $$ It now follows that
$$ \begin{split} \frac{Z^T\Sigma^{-1}Z}{Z^T(W+\Sigma_0d_0)^{-1}Z} &= \frac{W+\Sigma_0d_0}{\Sigma} \\ &= \frac{W}{\Sigma} + \Sigma_0d_0\Sigma^{-1} \\ &\sim \chi^2(d+d_0) \end{split} $$
since $\frac{W}{\Sigma}\sim \chi^2(d)$ and is independent of $\Sigma_0d_0\Sigma^{-1} \sim \chi^2(d_0)$
Special case 2: "$d_0 = 0$" If $V \sim \operatorname{Wishart}(\Sigma, d)$ then $$ \frac{1}{(V^{-1})_{ii}} \sim \frac{1}{(\Sigma^{-1})_{ii}}\chi^2(d-p+1) $$ see Inverse-Wishart.
Now for any fixed vector $a$ of length $p$. Take an orthogonal matrix $B=\{b_{(1)},\ldots ,b_{(p-1)},\frac{a}{|a|}\}$. Then $B^TV^{-1}B \sim \text{Wishart}(B^T\Sigma^{-1}B,d)$. We now have $$ \begin{split} (B^TVB)^{-1} &= B^{-1}V^{-1}B^{T-1} = B^TV^{-1}B \\ (B^T\Sigma B)^{-1} &= B^T\Sigma^{-1}B \\ (B^T\Sigma^{-1}B)_{pp} &= \frac{1}{|a|^2}a^T\Sigma^{-1}a \\ (B^TV^{-1}B)_{pp} &= \frac{1}{|a|^2}a^TV^{-1}a \end{split} $$ Using the above result we have that for any $a$ $$ \frac{Z^T\Sigma^{-1}Z}{Z^TW^{-1}Z} \mid Z = a \sim \chi^2(d-p+1). $$ Since the conditional distribution does not depend on $a$ it holds unconditionally as well.