No, uintptr_t is an integer type to which any object pointer type can be converted without loss of information. (Strictly speaking, the guarantee is for conversion to and from void*.) And if an implementation doesn't have a sufficiently wide integer type, it won't define uintptr_t. (Likewise for intptr_t the signed equivalent.)

There's no guarantee that a function pointer type can be converted to uintptr_t without loss of information.

C currently has two kinds of pointer types: object pointer types and function pointer types. "Fat pointers" could be a third. And since a fat pointer would internally be similar to a structure, converting it to or from an integer doesn't make a whole lot of sense. (If you want to examine the representation, you can use memcpy to copy it to an array of unsigned char.)