The result is true, as stated. Changing slightly the notation we have to prove that the Fourier transform of $x^\beta |x|^{-s} \phi$ is in $L^1(\mathbb R^n)$ for $\phi \in C_0^\infty (\mathbb R^n)$, $\phi=1$ in a neighborhhod of 0.

The function $h(x)=x^\beta |x|^{-s}$ is homogenuous of degree $0<|\beta|-s<1$. Integrating by parts $${\cal F}(h\phi)(\xi)=\frac{c}{\xi_j}\cal F(D_j (h\phi))=\frac{c}{\xi_j}\cal F(\phi D_j h)+...$$ where the dots indicate terms which decay faster then polynomials as $|\xi| \to \infty$ (the function $hD_j \phi$ is smooth).

Let us now consider the Fourier transform of $\phi D_j h$. Since $D_j h$ is of the form $a_1x^{\gamma_1} |x|^{-s}+x^{\gamma_2}|x|^{-s-2}$ with $|\gamma_1|=|\beta|-1$, $|\gamma_2|=|\beta|+1$ we can use Stein-Weiss "Introduction to Fourier Analysis in Euclidean spaces" Theorem 4.1, Chapter 4 pag. 160 to get for large $|\xi|$
$$| {\cal F}(D_j h(\xi)| \leq \frac{C}{|\xi|^{n+|\beta|-s-1}}.$$
Note that Theorem 4.1 requires negative exponents $\gamma_1-s$, etc. and this is the reason for integrating by parts. Also it is stated for spherical harmonics but applies to homogenuous polynomials, by Theorem 2.1 p. 139. The Fourier transform of $\phi D_j h$ differs from that of $D_j h$ by a fast decaying function (see (*) below) and has the same decay. Finally
$$| {\cal F}(h\phi )| \leq \frac{C}{|\xi|^{n+|\beta|-s}}$$ which gives integrability at infinity.

(*) EDIT $\phi D_j h-D_j h=(1-\phi)D_j h$ is smooth but not integrable at infinity. Its (distributional) Fourier transform $g$ is a function, being the difference of the Fourier transforms of $\phi D_j h$ and $D_j h$. Fix $N \in \mathbb N$; then $|\xi|^{2N} g$ is the Fourier transform of $\Delta^N [(1-\phi)D_j h]$ and, distributing derivative and using that $D^\alpha \phi$ has compat support for $|\alpha| \geq 1$, it follows that $\Delta^N [(1-\phi)D_j h] \in L^1(\mathbb R^n)$ if $N$ is sufficiently large. Then $|\xi|^{2N}g$ is bounded. This argument is from the book of Grafakos, classical Fourier analysis, Example 2.4.9 pag 142.

Another possibility (which simplifies some parts) is to differentiate n-times instead of 1 at the beginning of the proof then getting ah homogenuous function $D^\alpha h$, $|\alpha|=n$, of degree $-n<|\beta|-s-n<-n+1$. If $n \geq 3$ then $D^\alpha h \in L^1+L^2$ and its Fourier transform is a function which is homogenuous of degree $s-|\beta|$ (but local boundedness needs an argument). The last part concerning the decay of the Fourier transform of $(1-\phi)D^\alpha h$ is easier since, using the definition, one can integrate by parts on large balls $B_R$ and the boundary terms tend to 0 as $R \to \infty$.