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 $\|b^*\|^2_2 = h(bb^*) = h(b^*b) = \|b\|^2_2$.
(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. 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"...)