When you say $r(A)$ is almost linear, do you mean that for large $A$ that $r(A)\approx\rho A+c$ for some constants $\rho$ and $c$? Also, what do you mean by $\phi(A)\rightarrow\pm\frac{\pi}{2}$?
Without the $\pm$, and with the above assumption about $r(A)$:
\frac{d}{dA}\Gamma(0,r(A)e^{i\phi(A)})=-\left[r(A)e^{i\phi(A)}\right]^{-1}e^{-\left[r(A)e^{i\phi(A)}\right]}\left(r'(A)e^{i\phi(A)}+r(A)i\phi'(A)e^{i\phi(A)}\right)
For large $A$, $r(A)\approx\rho A+c$, r'(A)\approx\rho, $\phi(A)\approx\frac{\pi}{2}$, and \phi'(A)\approx0. And the above reduces to:
$\frac{d}{dA}\Gamma(0,r(A)e^{i\phi(A)})\approx-\left[(\rho A+c) i\right]^{-1}e^{-\left[(\rho A+c)i\right]}\left(\rho i +0\right)$
$\frac{d}{dA}\Gamma(0,r(A)e^{i\phi(A)})\approx-\frac{1}{A+\frac{c}{\rho}}e^{-(\rho A+c)i}$
So for large $A$, increasing $A$ by a differential $\Delta A$ will "add" about $\frac{\Delta A}{A+\frac{c}{\rho}}$ to your quantity, but in the direction of $-e^{-(\rho A+c)i}$; that is it will add a complex differential of about $\frac{\Delta A}{A+\frac{c}{\rho}}$ in magnitude at an angle of $-\rho A-c+\pi$.
Imagine an angle that spins around the origin (clockwise if $\rho$ is positive) at a linear rate with respect to $A$. Now add to your quantity in that rotating direction with a magnitudes that decrease roughly like $\frac{1}{A}$.