I think that the earlier answers haven't adequately explained why there isn't a force law for the strong force. It's not because it's relativistic or quantum (electromagnetism is both of those). It's because it's nonlinear.
In the gravitational and electromagnetic formulas in the question, $q_1$ and $q_2$ (or $m_1$ and $m_2$) are the charges at exactly two points, and $r$ is the distance between those two points. The formulas can't be used if there are more than two charges. You can extend them to the general case by turning them into sums over pairs of points, but that only makes sense if the force is linear.
You can write down a two-particle force law for the strong force. It's $F=-V'(r)$ where $V(r)$ is the function in Johannes's answer. What you can't do is use that same force law when there are more particles – not because the law is wrong as such, but because you can't use linearity to generalize it.
Though there is no simple force law, you can build an intuition for the magnitude and direction of the strong force from these rules:
Strong force field lines prefer to cluster in regions with a diameter of about 1 fm. They attract each other when farther apart and repel each other when closer (this is the nonlinearity).
If the total strong charge in an enclosed region of space isn't "zero" (color singlet), then there have to be strong-force field lines crossing the boundary of the region. (This is the strong-force version of Gauss's law.)
The energy density of the field is roughly constant where there are field lines and zero elsewhere.
The force is minus the gradient of the potential.
If there are two charges in an otherwise empty universe, then applying "Gauss's law" to any region containing one particle and not the other tells you that there have to be field lines crossing the space between the particles. The lines can follow any path, but the lowest-energy configuration is that in which they're all in a straight "flux tube" with a diameter of about 1 fm. In that configuration, the only way to reduce the field energy is to bring the charges closer together, so there is a force in that direction, i.e., of each particle directly toward the other. The energy density of the tube per unit distance is constant, so the strength of the force is independent of the length of the tube. (It's about 10,000 newtons; that's the $k$ in Johannes's answer.)
If there are three or more charges, no proper subset of which is color neutral, then again there have to be field lines connecting all of the particles. The shortest, hence lowest energy, way to connect them is a Steiner tree. The force on each charge is toward the nearest vertex of the Steiner tree, and the magnitude of it is again a constant 10,000 N if the separations are large enough.
This is only an approximation, but it's not too bad. In figure 4 of this paper, which shows pentaquarks simulated in lattice QCD, you can clearly see the Steiner-tree structure.
Note that it's very common to see diagrams with incorrect flux tube geometries. For example, all of the configurations in this illustration are wrong except the single tube in the $u\overline c$ meson.
This approximation ignores pair production of quarks. The practical effect of pair production is that the flux tubes connecting charges can't get very long, because it's energetically cheaper to create quarks which allow them to "break".
Quark-like particles with high rest masses, charged under a strong-like force, could be connected by much longer flux tubes. Such particles are known as "quirks" and there are a few papers about them, such as this one which proposes searching for them at the LHC.
I said that the energy density of the field is zero where there are no field lines, but that isn't quite true: in principle it's nonzero out to infinity, but it dies off exponentially.
"Gauss's law" says that there aren't (or at least needn't be) flux tubes connecting different hadrons, but there is a residual strong force between hadrons, which dies off exponentially with distance. It's weak enough that it can (as far as I know) be treated as linear, and there is a "pairwise" force law for it like the gravitational and electrostatic laws. That's $F=-\nabla V$ where $V$ is the function in dmckee's answer.