Your naïve interpretation of the work equation doesn't quite make sense in this context. Consider that the standard formula for work gives that $W=\int {\vec F}\cdot d{\vec x}$. At minimum, the presence of the dot product in the above equation should not be ignored, and we should interpret, for a radial force, the work to be equal to $\int \frac{F\,dr}{\sqrt{1-\frac{2M}{r}}}^{1}$. We should then note that this factor will exactly cancel the effect you cite.
${}^{1}$Note that, properly, we are defining the line integral in terms of three unit vectors normal to the line, so, the integral would have the form $\int \sqrt{|g|}\epsilon_{abcd}{\hat t^{a}}{\hat \theta^{b}}{\hat \phi^{c}}F^{d}$. When this is completely simplified, you'll find that the measure of the integral $\sqrt{|g|}$ has a factor of $\sqrt{1-\frac{2M}{r}}$ that cancels against the factor of $\frac{1}{1-\frac{2M}{r}}$ in $g_{rr}$, producing the term seen above.