Let $k=\int_{S^{n-1}}du=\dfrac{2\pi^{n/2}}{\Gamma(n/2)}$ and $\Pi=I-P_u=uu^T$. Then
\begin{align*}
\sum_{i<j}\lambda_i\lambda_j
&=\frac12\left[\left(\operatorname{tr}(M)\right)^2-\operatorname{tr}(M^2)\right],\\
\left(\operatorname{tr}(P_uBP_u)\right)^2
&=\left(\operatorname{tr}(P_uB)\right)^2\\
&=\left(\operatorname{tr}(B)-\operatorname{tr}(\Pi B)\right)^2\\
&=\left(\operatorname{tr}(B)\right)^2 -2\operatorname{tr}(B)\operatorname{tr}(\Pi B) + \left(\operatorname{tr}(\Pi B)\right)^2,\\
&=\left(\operatorname{tr}(B)\right)^2 -2\operatorname{tr}(B)\operatorname{tr}(\Pi B) + (u^TBu)^2\\
&=\operatorname{tr}(B)\,\operatorname{tr}\big((I-2\Pi)B\big) + (u^TBu)^2,\\
\operatorname{tr}\left((P_uBP_u)^2\right)
&=\operatorname{tr}(P_uBP_uB)\\
&=\operatorname{tr}\left((I-\Pi)B(I-\Pi)B\right)\\
&=\operatorname{tr}(B^2)-2\operatorname{tr}(\Pi B^2)+\operatorname{tr}(\Pi B\Pi B)\\
&=\operatorname{tr}(B^2)-2\operatorname{tr}(\Pi B^2)+(u^TBu)^2\\
&=\operatorname{tr}\big((I-2\Pi)B^2\big) + (u^TBu)^2,\\
\int_{S^{n-1}}\sigma_2(P_uBP_u)du
&=\int_{S^{n-1}}\frac12\left[\left(\operatorname{tr}(P_uBP_u)\right)^2-\operatorname{tr}\left((P_uBP_u)^2\right)\right]du\\
&=\frac12\int_{S^{n-1}}\left[\operatorname{tr}(B)\,\operatorname{tr}\big((I-2\Pi)B\big)
-\operatorname{tr}\big((I-2\Pi)B^2\big)\right]du\\
&=\frac12\left[\operatorname{tr}(B)\,\operatorname{tr}\left(\left(k-\frac{2k}{n}\right)B\right)
-\operatorname{tr}\left(\left(k-\frac{2k}{n}\right)B^2\right)\right]\\
&=\frac{k}{2}\left(1-\frac{2}{n}\right)\left[
\left(\operatorname{tr}(B)\right)^2
-\operatorname{tr}(B^2)\right]\\
&=k\left(1-\frac{2}{n}\right)\sigma_2(B).
\end{align*}
(Note: we do not assume that $\operatorname{tr}(B)=0$ in the above.) It follows that $\int_{S^{n-1}}\sigma_2(P_uBP_u)du=0$ when $n=1$ (which is evident, because $P_u=0$) or $n=2$ (which again is evident, because $P_uBP_u$ has exactly two eigenvalues and at least one of them is zero).
When $n\ge3$, $\int_{S^{n-1}}\sigma_2(P_uBP_u)du$ has the same sign as $\sigma_2(B)$. In this case, if $\operatorname{tr}(B)=0$, then the integral is negative when $B\ne0$ and zero otherwise.
