Path coloring in general graphs
Edit: In the end I did find a reference for the reduction I remembered. It is given in Edge intersection graphs of single bend paths on a grid by Golumbic, Lipshteyn, and Stern. So, to sum up, the class of graphs obtained as intersections of paths on a grid contains all graphs. Therefore, path coloring on grids is as hard to approximate as graph coloring ($n^{1-\epsilon}$-hard).

Original answer:
Path Coloring can be reduced to graph coloring in the obvious way (make a vertex for each path, add an edge if two paths intersect). If the underlying graph in path coloring is allowed to be a grid there is also a reduction in the opposite direction, which means that in this case the problems are equivalent. The reduction does not change the number of colors, so all hardness of approximation results for graph coloring carry over to this case of path coloring, and therefore also when the underlying graph can be any graph.

I heard this reduction in a talk a while ago, but I can't seem to be able to find a reference, so I'll sketch it here.

We start with a graph $G(V,E)$ on $n$ vertices and the question is if it can be colored with $k$ colors. Construct a graph $H$ which is a $n\times ({n\choose 2}+1)$ grid (we index the rows by the vertices of $G$ and the columns by pairs of vertices of $G$). We will add one path in this graph for each vertex of $G$. Initially, each of these paths is simply a row of the grid, so no two of them intersect and they all have length $n\choose 2$. The idea here is that we will use the $n\choose 2$ edges between columns of the grid to simulate the edges of the original graph. For example, if vertex $i$ is connected to vertex $j$ in $G$, with $i&lt;j$, take the path that runs through row $i$ and re-route it as follows: the path now goes from vertex $(i,(i,j))$ to $(j,(i,j))$, then it uses the next edge on the $j$-th row and then it goes back to row $i$. This creates an intersection between paths $i$ and $j$, but no other edge intersections. By adding the rest of the edges of $G$ this way, the resulting instance is $k$-colorable iff $G$ is.