Yes, $\widetilde{A}_r$ and $A_r$ are canonically isomorphic. This follows from the following result in which you could take as $(B,\Delta_B)$ the universal C$^*$-algebraic compact quantum group $(A_u,\Delta_u)$ associated with $(A_0,\Delta_0)$.
Proposition. Let $(B,\Delta_B)$ and $(A,\Delta_A)$ be C$^*$-algebraic compact quantum groups and let $\pi : B \to A$ be a surjective unital $*$-homomorphism satisfying $(\pi \otimes \pi) \circ \Delta_B = \Delta_A \circ \pi$. Let $B_0 \subset B$ be a dense Hopf $*$-subalgebra. Assume that the restriction $\pi|_{B_0}$ is injective. Denote by $h_B$ and $h_A$ the Haar states on $(B,\Delta_B)$ and $(A,\Delta_A)$, respectively. Denote by $\pi_B$ and $\pi_A$ the corresponding GNS-representations.
Then, $h_B = h_A \circ \pi$ and there is a unique $*$-isomorphism $\theta : \pi_B(B) \to \pi_A(A)$ satisfying $\theta \circ \pi_B = \pi_A \circ \theta$.
Proof. Since $\pi|_{B_0}$ is injective, the restriction of $h_A \circ \pi$ to $B_0$ is an invariant state on the Hopf $*$-algebra $B_0$. By uniqueness of the invariant state, $h_A \circ \pi$ and $h_B$ are equal on $B_0$. By density of $B_0 \subset B$, we get that $h_B = h_A \circ \pi$. Since $\pi$ is assumed to be surjective, the rest follows immediately.
