Could chromatic number be easy to calculate when colouring is hard for some graph class?
Yes. Since 3-COLORING is FNP-complete, for any problem in FNP we can construct a graph G that is 3-colorable if and only if the problem has a solution. So you can pick your favorite problem from TFNP\FP (if it's not empty!), like finding a factor of a composite number, or a second Hamiltonian-cycle in a 3-regular graph, and convert it into a graph.

In more detail with the example of finding a factor.
We will fix an algorithm that converts a composite number $N$ into a graph $G(N)$ such that from $G(N)$ you can decode the value of $N$.
Compositeness of $N$ is a necessary for $G(N)$ to be in our class.
Moreover, any proper $3$-coloring of $G(N)$ can be converted into a proper divisor $d$ of $N$.
This can be done, e.g., by writing up a CNF that describes whether a given $d$ encoded in binary as $x_1\ldots x_{n}$ divides a fixed $N$ (that also has $n$ digits in binary, but these are not variables).
From satisfiability of a CNF there is a standard reduction to 3-COLORABILITY such that you can easily convert a proper $3$-coloring back to an $x_i$ satisfying assignment.
Thus solving the coloring problem, you would find a divisor $d$.
Moreover, by adding some simple extra gadgets, you can also decide whether a graph is of this form or not.
Our class is formed by graphs created this way, which are thus all $3$-colorable.