Let $(X,\mathcal O_X)$ be a locally noetherian scheme of dimension $1$ and suppose that $X$ is regular, that is: $\mathcal O_{X,x}$ is a regular local ring. We have no other hypothesis on $X$.
What is in this case the function field $K(X)$? This object in general is defined for integral schemes, is it also defined for non-integral schemes?
The function field is only defined for integral schemes (for which it is the local ring of the generic point). A regular scheme (of any dimension) doesn't have to be integral (there exist connected affine schemes all of whose local rings are domains but which are not themselves integral, see http://stacks.math.columbia.edu/tag/0568). It's impossible to come up with an example of a connected regular scheme (affine or otherwise) which is not integral because a connected scheme with locally finite set of connected components (e.g. locally Noetherian) will be irreducible (and hence integral) if its local rings are domains. But certainly one can come up with trivial examples of $1$-dimensional affine regular schemes which are not connected: take a discrete valuation ring $R$ and look at a disjoint union of finitely many copies of $\mathrm{Spec}(R)$ (more than one copy, of course).
I suppose the substitute for the function field of an integral scheme on a general scheme $X$ would be the direct product of the residue fields of its generic points (this reduces of course to the function field if the scheme is integral). I know at least one book where, for Dedekind schemes, this construction is used, and is called the ring of rational functions on the scheme. Indeed, Dedekind schemes are examples of the schemes you're considering, as they are Noetherian of dimension $1$ and normal, hence regular, but not always required to be connected (equivalently irreducible). The irreducible components of such a scheme coincide with the connected components, each of which is an integral Dedekind scheme, and the ring of rational functions as I've just defined it is the product of the function fields of these irreducible components. (Actually it might be that some sources allow Dedekind schemes to be zero-dimensional, but I don't know how standard that is.)