I think the follwing is the answer of this question. It is not depend on the above comments because I didn't understand them. This is just my approach.
(But not fully my idea because it depends on a strong proposition which is already known, and the remaining part is just a corollary.)

Proposition. Let $f$ be a modular form of weight $k$, defined over $\mathbb Q^{\mathrm{alg}}$. If $f$ is non constant, then the functions $f, Df$ and $D^2 f$ are algebraically independent over the function field $\mathbb C(q)$.

The proof is in the book, 'Introduction to Algebraic Independence Theory', Y.V.Nesterenko and P.Philippon, (Eds), Ch.1 prop 1.1. Here $q=e^{2\pi i \tau}$ and $D=\frac{1}{2\pi i}{\tau}=q\frac{d}{dq}$.

Let $\Delta = E_4^3-E_6^2$, cusp form of weight 12. Due to Ramanujan, we have the following formulas :

$$
DP=\frac{1}{12}(P^2-Q), \quad DQ=\frac{1}{3}(PQ-R), \quad DR=\frac{1}{2}(PR-Q^2).
$$

Thus we have $D\Delta = P\Delta$. Take $D$ in both side and do some elementary algebra, we have

$$
Q=13\frac{(D\Delta)^2}{\Delta^2}-12\frac{D^2 \Delta}{\Delta}.
$$

By the proposition, $\{\Delta, D\Delta, D^2\Delta\}$ is algebraically independent over $\mathbb C(q)$, so $\{Q^3-R^2,P,Q\}$ is algebraically independent over $\mathbb C(q)$, so $\{P,Q,R^2\}$ is algebraically independent over $\mathbb C(q)$. This is equivalent to say that $R^2$ is transcendental over $\mathbb C(q)[P,Q]$, and this implies that $R$ is transcendental over $\mathbb C(q)[P,Q]$, i.e. $\{P,Q,R\}$ is algebraically independent over $\mathbb C(q)$.