Let us consider this qualitatively with crystal field theory (CFT). Co is a transition metal with $d^7$ configuration. CFT says that depending on the geometry of the metal complex, the d-orbital energies will split into different energy levels. If the ligands (in this case, the four $\ce{SCN-}$ ligands) align with certain orbitals, then the electrons in these orbitals will feel an increased repulsion due to the closeness of the ligands, and increase in energy. The orbitals in-between the aligned axes, will not feel this repulsion, and will not increase in energy. The splitting diagram might look something like this
![Splitting diagram for a tetrahedral complex](https://i.sstatic.net/H4eGH.jpg)
There is an equation that relates the (splitting)
energy to wavelength:
$$
\Delta E = \frac{hc}{\lambda}
$$
From this equation, we see that if the splitting energy increases, wavelength will have to decrease, i.e. be shorter than before. This means a shift towards the violet region. As you say, $\ce{SCN-}$ is a strong ligand, meaning it leads to a rather high $\Delta E$. So we could say that $\ce{Co(SCN)_4^{2-}}$ is blue due to the high energy gap between the d-orbitals. Electrons are excited from one of the lower orbitals up to one of the higher orbitals. Upon de-exciting, light is emitted with an energy equal to $\frac{hc}{\lambda}$. This happens to be in the blue/violet region.
Weaker ligands would split the energy levels less, and the complex's color would therefore have a color closer to the red part of the visible spectrum.
Did this make it a little clearer?