If, for real positive definite symmetric $A, B, C$,
$$\log\det (A+B) \geq \log\det(A+C)$$
then can it be said that
$$\log\det(B) \geq \log\det(C)?$$
NOTE: A crude form of the reverse is certainly true. That is, if: $$B \succeq C$$ then $$\log\det(A+B) \geq \log\det(A+C)$$ However, this does not help with the forward problem.
No, not even for diagonal matrices. Consider for instance $$C = \left[\begin{array}{cc}\sqrt{2}-1 & 0 \\ 0 & \sqrt{2}-1\end{array}\right]$$ $$B = \left[\begin{array}{cc}1 & 0\\0 & \epsilon\end{array}\right]$$ $$A = I.$$
Then $$\det(A+B) = 2+2\epsilon \geq 2 = \det(A+C)$$ but $$\det(B) = \epsilon < 3-2\sqrt{2} = \det(C)$$ for sufficiently small $\epsilon.$