Yes, it is true pretty much for the reason you're saying, but you don't need Hodge theory or even to work over $\mathbb C$. 

Using the notations of the question, in addition let $Y=\mathrm{Spec}(S/I)$ be the affine cone over $X$, $P\in Y$ the vertex, and $U=Y\setminus \{P\}$. 
Finally, let $\mathrm{depth}(S/I)=d\geq 3$.
First of all we have a long exact sequence:

$$
\dots \to H^i(Y,\mathscr O_Y) \to H^i(U,\mathscr O_U) \to H^{i+1}_P(Y,\mathscr O_Y) \to H^{i+1}(Y,\mathscr O_Y)  \to  \dots.
$$
Since $Y$ is affine, this implies that for $i>0$,
$$
H^i(U,\mathscr O_U) \simeq H^{i+1}_P(Y,\mathscr O_Y) 
$$
and hence 
$$
H^i(U,\mathscr O_U)=0
$$
for $0< i < d-1$.

Now the desired vanishing (and more) follows from the following:

> **Proposition**
$\quad\  
H^i(U,\mathscr O_U) \simeq \bigoplus_{n\in\mathbb Z} H^i(X, \mathscr O_X(n)) 
$  

**Proof** $U$ is an $\mathbb A^1$-bundle over $X$. In fact, it is easy to see that $U\simeq \mathrm{Spec}_X ( \oplus _{n\in \mathbb Z} \mathscr O_X(n))$ with a projection $\pi:U\to X$. It follows that $\pi_*\mathscr O_U\simeq \oplus _{n\in \mathbb Z} \mathscr O_X(n)$ and $R^j\pi_*\mathscr O_U=0$ for $j>0$. Then the claimed isomorphism follows from the simple special case of the Leray spectral sequence when all $R^j$'s with $j>0$ are $0$.