The case $p=\infty$ is easy: let $M=\|f-g\|_\infty$. Since $f\le g+M$, it follows that $f^*\le g^*+M$, and we are done.
For $1\le p<\infty$ the function $t\mapsto |t|^p$ is convex on $\mathbb R$, which means that we can use Theorem 3.5 on the very same page of the book you are reading.
3.5 Theorem (nonexpansivity of rearrangement). Let $J:\mathbb R\to\mathbb R$ be a nonnegative convex function such that $J(0)=0$. Let $f$ and $g$ be nonnegative functions on $\mathbb R^n$, vanishing at infinity. Then
$$\int_{\mathbb R^n} J(f^*-g^*)\le \int_{\mathbb R^n} J(f-g)\tag{1}$$
If we also assume that $J$ is strictly convex, that $f=f^*$, and that $f$ is strictly decreasing [in radial direction], then equality in (1) implies that $g=g^*$.
I suspect you asked this question because of the sentence "the obvious generalization is $\|f^*-g^*\|_p \le \|f-g\|_p$". I guess the authors did not mean to say "this generalization is obviously true", but rather "obviously, this is a generalization".