The correct answer is $\boxed{\frac{\pi}{2}}$.
The integral of a function $f(x, y)$ over a path $C$ is given by
$$\int_C f(x, y) \, ds = \int_a^b f(x(t), y(t)) |x'(t) + y'(t)| \, dt$$
where $(x(t), y(t))$ is a parametrization of $C$.
In this case, the path $C$ is the quarter-circle $x^2 + y^2 = 1$, $y \geq 0$, traversed in a counterclockwise sense. A parametrization of this path is
$$x(t) = \cos t, \quad y(t) = \sin t, \quad 0 \leq t \leq \frac{\pi}{2}$$
Substituting this into the formula for the integral, we get
$$\int_C (x + y)^2 \, ds = \int_0^{\frac{\pi}{2}} (\cos t + \sin t)^2 |-\sin t + \cos t| \, dt = \int_0^{\frac{\pi}{2}} 2 \cos^2 t + 2 \sin^2 t \, dt = \int_0^{\frac{\pi}{2}} 1 \, dt = \frac{\pi}{2}$$
Therefore, the integral of $(x + y)^2$ on path $AB$ traversed in a counterclockwise sense is $\boxed{\frac{\pi}{2}}$.
The other options are incorrect because they do not take into account the correct parametrization of the path $C$.