I think there are some subtle points here about what the "right action" even means.

For a general $*$-algebra $A_0$ and a functional $\phi:A_0\rightarrow\mathbb C$, we first of all have to decide what "positive" means for $\phi$. We could take this as being $\phi(a^*a)\geq0$ for all $a$. Then Cauchy-Schwarz holds and we can form $L^2(A_0,\phi)$. Why, however, do we have that left multiplication $\pi_L:A_0\rightarrow L(A_0), \pi_L(a)(b) = ab$ extends to a *bounded* operator on $L^2(A_0,\phi)$? For $C^*$-algebras, this is a basic but slightly subtle result: it follows from the inequality $b^*a^*ab \leq \|a\|^2 b^*b$.

For compact quantum group algebras $A_0$ I think you have to use the unitary corepresentation theory to show that $\pi_L(a)$ is a bounded operator, for each $a\in A_0$. Indeed, this is so: see Section 5.4.2 of Timmermann's book.

Thus, in general, there really is no notion of "right action", because again why need $\pi_R(a):b\mapsto ba; A_0\rightarrow A_0$ extends to a bounded operator on $L^2(A_0)$? Even for $C^*$-algebras, you do not have a right action. However, if $h$ is a *trace* (corresponding to $A$ being of *Kac type*) then a simple calculation shows that $\pi_R$ exists and gives the same norm:
$$ \|\pi_R(a)(b)\|^2_2 = h(a^*b^*ba) = h(baa^*b^*) = \|\pi_L(a^*)(b^*)\|^2_2
\leq \|\pi_L(a^*)\|^2 \|b^*\|^2_2 $$
and $\|b^*\|^2_2 = h(bb^*) = h(b^*b) = \|b\|^2_2$.

Actually, for CQG algebras we can say more, because the state $h$ is KMS. This is not so clear in Timmermann's book I think, but compare Theorem 8.1.13 (ii) with Example 8.1.22. In particular, see the top of page 213. In short, there is an automorphism (not a $*$-automorphism) $\sigma_{i/2}$ of $A_0$ with
$$ h(a^*a) = h(\sigma_{i/2}(a) \sigma_{i/2}(a)^*) \qquad (a\in A_0). $$
Then consider
$$ \| \pi_R(a)(b)\|^2_2 = h(a^*b^*ba) = h(\sigma_{i/2}(ba)\sigma_{i/2}(ba)^*)
= h(\sigma_{i/2}(b) \sigma_{i/2}(a) \sigma_{i/2}(a)^* \sigma_{i/2}(b)^*)
= \|\pi_L(\sigma_{i/2}(a)^*)(\sigma_{i/2}(b)^*)\|_2^2, $$
and note that $\|\sigma_{i/2}(b)^*\|_2^2 = h(\sigma_{i/2}(b)\sigma_{i/2}(b)^*)
= h(b^*b) = \|b\|^2_2$, here for $a,b\in A_0$. It follows that $\pi_R(a)$ is bounded with $\| \pi_R(a) \| \leq \|\pi_L(\sigma_{i/2}(a)^*)\|$ (actually, equal). So, $\pi_R$ is well-defined, but it does not induce the same norm, except when $A$ is Kac.

(My motivation for this argument was the following: One possible way to give some notion of a right action comes from von Neumann algebra theory, and the theory of "correspondences", see Takesaki, Volume 2, Chapter IX, Section 3. As $h$ is a KMS state, $\pi_L(A_0)''$ is in standard position on $L^2(A_0)$, and so there is a modular operator $J$. We can then *define* $\pi_R(a) = J\pi_L(a)^*J$. By definition, this gives the same norm on $A_0$, but it's not really what you would call the "right multiplication"...)