Your conjecture is correct: The rise of a liquid in a capillary is not just a function of the liquid-air surface tension but also the liquid-solid surface energy, AND this liquid-solid surface energy is present in the equation and its effect is represented by the contact angle parameter in the capillary rise equation.
Derivation of the capillary rise equation appears to be a bit involved, but there seems to be a good description here: Capillary Rise Equation Derivation
Note Figure 8.1 in the linked document: Even though it is the same liquid and same air for the two capillaries shown, in one capillary the rise is positive while in the other the rise is negative. The difference between the two capillaries? In one the contact angle is positive while in the other the contact angle is negative, presumably because the two capillaries are made of different materials so they have different liquid-solid surface energies.
We have to make a few assumptions here. The first would be that the soap solution mixes with the water evenly and instantaneously in the capillary - this would be a good approximation, as it's a fairly small volume in the tube, on a similar order of magnitude to the volume of the drop. The next is the form of the meniscus:
Please excuse my crude diagram! You can see that the meniscus is approximated to have a flat bottom, with the sloped sides making an angle $\theta$ - this is symmetrical. The difference in the liquid height from the surface of the body to the meniscus will be $h$. The capillary will have a radius of $r$.
We can take the difference in the surface free energy by changing the height as
$2\pi rh(\gamma_{sl}-\gamma_{sg})$;
$\gamma_{sl}$ is the surface free energy of the solid-liquid interface, and $\gamma_{sg}$ is the surface free energy of the solid-gas interface.
Then, we also take the difference in the gravitational potential energy in changing the height of the liquid:
$\pi r^2h\rho\times\frac{h}{2}g$,
where $\rho$ is the density of the liquid, and we take the centre of mass of the liquid in the capillary to be at $h/2$.
So, the free energy of the meniscus is given by
$\frac{\pi}{2}r^2h^2\rho g+2\pi rh(\gamma_{sl}-\gamma_{sg})$.
We have a stable meniscus when this is minimised, so we differentiate it with respect to $h$, and set that equal to zero, giving
$h=\frac{-2(\gamma_{sl}-\gamma_{sg})}{\rho gr}$.
We can then substitute in the Young Equation, finally giving
$h=\frac{2\gamma_{lg}cos\theta}{\rho gr}$.
The surface free energy of the liquid-gas interface, $\gamma_{lg}$, will be determined by the soap, which would also alter $\rho$. Also, note that $\theta$ can be negative - in this case the level in the capillary would fall, which would be the case with soap using the numbers you gave.
With these assumptions, the water level would change near instantaneously - though it may require some mixing.
Best Answer
Think of a barometer.
The hydrostatic pressure can not exceed the difference between the atmospheric pressure and vapor pressure.
$101325Pa -2000Pa = \rho gh$
Then use $h$ in your equation to find $R$.