The answer is yes always, provided $f\colon\thinspace (X,x_0)\to (Y,y_0)$ is a *pointed* homotopy equivalence of pointed spaces (meaning that the homotopies $g\circ f \simeq 1_X$ and $f\circ g\simeq 1_Y$ preserve the base points).

This follows from the fact that $\Omega$ is a homotopy functor from based spaces to based spaces, meaning in particular that

$$\Omega(g)\circ\Omega(f) = \Omega(g\circ f) \simeq \Omega(1_X) = 1_{\Omega X}$$

and

$$\Omega(f)\circ\Omega(g) = \Omega(f\circ g) \simeq \Omega(1_Y) = 1_{\Omega Y}.$$

To see that $\Omega$ is a homotopy functor, note that $\Omega(f)$ takes a loop $\gamma\colon\thinspace I\to X$ to the composition $f\circ \gamma\colon\thinspace I\to Y$. So if $F_t\colon X\to Y$ is a pointed homotopy from $f$ to $f'$, then
$\Omega(F_t)\colon\thinspace \Omega(X)\to \Omega(Y)$ is a pointed homotopy from $\Omega(f)$ to $\Omega(f')$.