**Short answer: yes.**

Let $\psi_\epsilon(x):=\frac{1}{\epsilon^n}\exp{\epsilon^2/(\epsilon^2-|x|^2)}$ for $|x|<\epsilon$, and $\psi_{\epsilon}(x)=0$ for $|x|\geq \epsilon$. Set $\epsilon=2$, and define
$\phi$ is the convolution of $C\phi_{\epsilon}$ with the characteristic function of $B_{r+3/2}(0)$, that is,

**$\phi(x):= C\psi_\epsilon(x)* \chi_{B_{r+3/2}(0)}$.** Here $C$ is a normalizing constant (this may not be needed, but I haven't checked).

This yields a smooth cut-off function which is 1 in the ball $B_{r+1}(0)$, and zero outside $B_{r+2}(x)$.

To see this does the trick, one can use a localization theorem, for example, Theorem 3.20 in 'Strongly Elliptic Systems and Boundary Integral Equations' by W. McLean. This theorem states:

'Suppose that $\phi \in C^r_{comp}(\mathbb{R}^n)$ for some integer $r\geq 1$, and let $|s|\leq r$. If $u\in H^s(\Omega)$ then $\phi u \in H^s(\Omega)$, and $||\phi u||_{H^s(\Omega)} \leq C_r||\phi||_{W^{r,\infty}(\mathbb{R}^n)}Q_u$ where $Q_u=||u||_{H^s(\Omega).
}$ *(Apologies, I encountered trouble while trying to typeset the LaTeX here).*

The same result holds with $H^s(\Omega)$ replaced with $\tilde{H}^s(\Omega)$.'

The proof proceeds using $\Omega = A_2$, and then

either by
(a) considering the situation for $s=r$, using duality to see it holds for $s=-r$, and the intermediate $s$ by interpolation. This is suggested by Yakov above.

or (b) by examining $\hat{\phi u}$ and using Peetre's inequality.

Since the constructed $\phi \in C^\infty$ and has compact support, it will satisfy the inequality you seek. In my comment I asked whether you wanted a $\phi$ of minimal regularity (relative to $\tau$); my construction works but may be overkill.