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There is no relationship between the two, even in the simplest case where the $n\times n$ matrices $A$ and $B$ are simultaneously diagonalizable. In that case, we can write the expressions in terms of the eigenvalues $\lambda^{(A)}_{j}$ and $\lambda^{(B)}_{j}$, of which there are $n$ each (counting with multiplicity). $${\rm tr}\left(A^{-1}B\right)=\sum_{j=1}^{n}\frac{\lambda^{(B)}_{j}}{\lambda^{(A)}_{j}}.$$ In contrast, the ratio of the traces is $$\frac{{\rm tr}\,B}{{\rm tr}\,A}=\frac{\sum_{j=1}^{n}\lambda^{(B)}_{j}}{\sum_{j=1}^{n}\lambda^{(A)}_{j}}.$$ Note that by taking one of the $\lambda_{j}^{(A)}$ to $0$, the magnitude ${\rm tr}\left(A^{-1}B\right)$ can be made arbitrarily large (and the sign either positive or negative if the eigenvalues are not restricted to be nonnegative), while the other expression is minimally affected.