$\newcommand\E{\mathsf E}$You want to find $$E:=\E\frac{X^T BX}{X^T AX},$$ where $X$ is a random vector uniformly distributed on $S^2$, so that $X$ equals $$\frac{[G_1,G_2,G_3]^T}{\sqrt{[G_1,G_2,G_3]^T[G_1,G_2,G_3]}}$$ in distribution, where $G_1,G_2,G_3$ are independent standard normal random variables.
Also, by the spherical symmetry, without loss of generality the matrix $A$ is diagonal. So, $$E=\E\frac{\sum_{i,j=1}^3 b_{i,j}G_iG_j}{\sum_{i=1}^3 a_i G_i^2},$$ for for some real $b_{i,j}$'s and some positive real $a_i$'s, $$E=\E\frac{\sum_{j,k=1}^3 b_{j,k}G_jG_k}{\sum_{i=1}^3 a_i G_i^2}=\sum_{j,k=1}^3 b_{j,k}\E R_{j,k},$$ where $$R_{j,k}:=\frac{G_jG_k}{\sum_{i=1}^3 a_i G_i^2}.$$ Note that $R_{j,k}$ will turn into $-R_{j,k}$ if $j\ne k$ and $G_j$ is replaced by $-G_j$. So, by symmetryif $j\ne k$, then the distribution of $R_{j,k}$ is symmetric and hence $\E R_{j,k}=0$, $$E=\E\frac{\sum_{j=1}^3 b_jG_j^2}{\sum_{i=1}^3 a_i G_i^2} =\sum_{j=1}^3 b_jE_j,$$So, $$E=\sum_{j=1}^3 b_jE_j,$$ where $b_j:=b_{j,j}$ and $$E_j:=\E\frac{G_j^2}{\sum_{i=1}^3 a_i G_i^2}.$$$$E_j:=\E R_{j,j}=\E\frac{G_j^2}{\sum_{i=1}^3 a_i G_i^2}.$$
So, the problem is equivalent to computing $E_j$. However, it is highly unlikely that $E_j$ can be expressed in terms of elementary or even special functions, even for specific values of the $a_i$'s. For instance, Mathematica cannot do anything for $E_j$ even when $j=1$, $a_1=1$, $a_2=2$, and $a_3=3$:
However, we have this:
