Resistance is a property of a component with linear relationship between current density $\mathbf{J}$ and electric field $\mathbf{E}$. In such cases one can show that there will be linear relationship between the voltage drop ($V$) between two points on that component and the current ($I$) flowing between them.
The general equations that govern this situation are (isotropic case):
$$
\begin{align}
\mathbf{J}=\sigma\mathbf{E}&=-\sigma\boldsymbol{\nabla}\phi \\
\nabla^2\phi &= 0\\
\end{align}
$$
Where $\phi$ is the electrostatic potential. You would solve them and then get relationship between voltage and current. Below is an example of a simple solution. In general solution may not be that simple, but it will give you what you are after - linear coefficient that relates voltage and current.
Simple solution for a cylindrical segment of a conductor
The resistance equation you have quoted is a simple solution of the equation I gave above.
In your case you could arrive at it by choosing a cylinder with axis of symmetry along z-axis, cross-section area $A=\pi R^2$, height $l$ and its centre at the origin. We choose the boundary conditions to be $\phi=0$ at the bottom face of the cyllinder $-\phi=V$ at the top face of the cylinder and $\partial_r\phi=0$ at the sides of the cylinder (no current electric field perpendicular to cyllinder walls).
One then finds, in cyllindrical coordinates with symmetry
$$
\nabla^2\phi=\frac{1}{r}\partial_{r}\left(r\partial_r\phi\right)+\partial_{zz}\phi=0
$$
Assuming a more general $\partial_r\phi=0$ which means no current is flowing side-ways, only along the length of the cyllinder, we get the solution $\phi=-\frac{V}{l}\cdot \left(z-l/2\right)$.
Current ($I$) is then the current density, $\mathbf{J}=-\sigma\boldsymbol{\nabla}\phi=\mathbf{\hat{z}}\sigma\cdot V/l$, flowing through the cross-section of the cyllinder, so $I=\pi R^2 \mathbf{\hat{z}}.\mathbf{J}=A\sigma \cdot V/l$. Finally, introducing $\rho=1/\sigma$:
$$
\frac{V}{I}=l\rho/A
$$
