Should useless type qualifiers on return types be used, for clarity?

It's usually better for your code to describe as accurately as possible what's going on. You're getting this warning because the const in const int foo(); is basically meaningless. The API only seems clearer if you don't know what the const keyword means. Don't overload meaning like that; static is bad enough as it is, and there's no reason to add the potential for more confusion.

const char * means something different than const int does, which is why your tool doesn't complain about it. The former is a pointer to a constant string, meaning any code calling the function returning that type shouldn't try to modify the contents of the string (it might be in ROM for example). In the latter case, the system has no way to enforce that you not make changes to the returned int, so the qualifier is meaningless. A closer parallel to the return types would be:

const int foo();
char * const foo2();

which will both cause your static analysis to give the warning - adding a const qualifier to a return value is a meaningless operation. It only makes sense when you have a a reference parameter (or return type), like your const char * example.

In fact, I just made a little test program, and GCC even explicitly warns about this problem:

test.c:6: warning: type qualifiers ignored on function return type

So it's not just your static analysis program that's complaining.

You can use a different technique to illustrate your intent without making the tools unhappy.

#define CONST_RETURN

CONST_RETURN int foo();

You don't have a problem with const char * because that's declaring a pointer to constant chars, not a constant pointer.

Ignoring the const for now, foo() returns a value. You can do

int x = foo();

and assign the value returned by foo() to the variable x, in much the same way you can do

int x = 42;

to assign the value 42 to variable x.
But you cannot change the 42 ... or the value returned by foo(). Saying that the value returned from foo() cannot be changed, by applying the const keyword to the type of foo() accomplishes nothing.

Values cannot be const (or restrict, or volatile). Only objects can have type qualifiers.

Contrast with

const char *foo();

In this case, foo() returns a pointer to an object. The object pointed to by the value returned can be qualified const.

The int is returned by copy. It may be a copy of a const, but when it is assigned to something else, that something by virtue of the fact that it was assignable, cannot by definition be a const.

The keyword const has specific semantics within the language, whereas here you are misusing it as essentially a comment. Rather than adding clarity, it rather suggests a misunderstanding of the language semantics.

const int foo() is very different from const char* foo(). const char* foo() returns an array (usually a string) whose content is not allowed to change. Think about the difference between:

 const char* a = "Hello World";

and

const int b = 1;

a is still a variable and can be assigned to other strings that can't change whereas b is not a variable. So

const char* foo();
const char* a = "Hello World\n";
a = foo();

is allowed but

const int bar();
const int b = 0;
b = bar();

is not allowed, even with the const declaration of bar().