How is callCC implemented in strict functional languages?

Consider the following example in Haskell of a function quux along with the definitions of the continuation monad and callCC.

instance Monad (Cont r) where
    return x = cont ($ x)
    s >>= f  = cont $ \c -> runCont s $ \x -> runCont (f x) c

callCC :: ((a -> Cont r b) -> Cont r a) -> Cont r a
callCC f = cont $ \h -> runCont (f (\a -> cont $ \_ -> h a)) h

quux :: Cont r Int
quux = callCC $ \k -> do
    let n = 5
    k n
    return 25

As I understand this example. The do block can be thought of as

k n >>= \_ -> return 25 == 
cont $ \c -> runCont (k n) $ \x -> runCont ((\_ -> return 25) x) c

And we can see from the definition of k which is \a -> cont $ \_ -> h a that in the above we have \x -> runCont ((\_ -> return 25) x) c being passed into the argument that is ignored with underscore. Ultimately the return 25 is effectively "ignored" because the underscore argument is never used so from lazy evaluation its never evaluated.

So as far as I can tell this implementation of callCC strongly fundamentally depends on lazy evaluation. How would this callCC be done in a strict (non-lazy) functional language?

No. This implementation of callcc doesn't depend upon lazy evaluation. To prove this I'll implement it in a strict functional language and show that anything after k n is not executed at all.

The strict functional language I'll be using is JavaScript. Since JavaScript is not statically typed you don't need to declare a newtype. Hence we start by defining the return and >>= functions of the Cont monad in JavaScript. We'll call these functions unit and bind respectively:

function unit(a) {
    return function (k) {
        return k(a);
    };
}

function bind(m, k) {
    return function (c) {
        return m(function (a) {
            return k(a)(c);
        });
    };
}

Next we define callcc as follows:

function callcc(f) {
    return function (c) {
        return f(function (a) {
            return function () {
                return c(a);
            };
        })(c);
    };
}

Now we can define quux as follows:

var quux = callcc(function (k) {
    var n = 5;

    return bind(k(n), function () {
        alert("Hello World!");
        return unit(25);
    });
});

Note that I added an alert inside the second argument to bind to test whether or not it's executed. Now if you call quux(alert) it will display 5 but it won't display "Hello World!". This proves that the second argument to bind was never executed. See the demo for yourself.

Why does this happen? Let's start backwards from quux(alert). By beta reduction it's equivalent to:

(function (k) {
    var n = 5;

    return bind(k(n), function () {
        alert("Hello World!");
        return unit(25);
    });
})(function (a) {
    return function () {
        alert(a);
    };
})(alert);

By beta reducing it again it becomes:

bind(function () {
    alert(5);
}, function () {
    alert("Hello World!");
    return unit(25);
})(alert);

Next by the beta reduction of bind we get:

(function (c) {
    return (function () {
        alert(5);
    })(function (a) {
        return (function () {
            alert("Hello World!");
            return unit(25);
        })(a)(c);
    });
})(alert);

Now we can see why "Hello World!" was never displayed. By beta reduction we're executing function () { alert(5); }. It's the job of this function to call its argument, but it never does. Because of this execution stops and "Hello World!" is never displayed. In conclusion:

The callcc function doesn't depend upon lazy evaluation.

The function created by callcc terminates after k is called not because of lazy evaluation but because calling k breaks the chain by not calling it's first argument and hence returns immediately.

This brings me back to your question:

And we can see from the definition of k which is \a -> cont $ \_ -> h a that in the above we have \x -> runCont ((\_ -> return 25) x) c being passed into the argument that is ignored with underscore. Ultimately the return 25 is effectively "ignored" because the underscore argument is never used so from lazy evaluation its never evaluated.

You're mistaken. As you can see k is (\a -> cont $ \_ -> h a) and the function (\x -> runCont ((\_ -> return 25) x) c) is passed into the the argument that's ignored by k. You were correct until then. However this doesn't mean that return 25 is not evaluated because of lazy evaluation. It's simply not evaluated because the function (\x -> runCont ((\_ -> return 25) x) c) is never called. I hope that cleared things up.