There are two kinds of primes $\mathfrak q \subset R[T]$. The first possibility is that $\mathfrak q = \mathfrak p[T]$ with $\mathfrak p = R \cap \mathfrak q$. Let us call such a prime "small". The second possibility is that the inclusion $\mathfrak p[T] \subset \mathfrak q$ is strict. Let us call such a prime "big".
Suppose we have a sequence of primes $\mathfrak q_0 \subset \mathfrak q_1 \subset \ldots \subset\mathfrak q_r$ in $R[T]$. Then we get a sequence of big/small. If the sequence has only one switch from small to big, then of course $r \leq \dim(R) + 1$. The problem comes from a sequence with multiple switches. But thinking about it for a moment we see that it suffices to prove the following.
Scholium: If $\mathfrak q_0 \subset \mathfrak q_1$ in $R[T]$ lies over $\mathfrak p_0 \subset \mathfrak p_1$ in $R$ and if $\mathfrak q_0$ is big and $\mathfrak q_1$ is small, then there is a prime strictly in between $\mathfrak p_0$ and $\mathfrak p_1$.
To prove this we argue by contradiction and assume there is no prime strictly in between. Observe that in any case $\mathfrak p_0 \not = \mathfrak p_1$ by our definition of big and small. After replacing $R$ by $(R/\mathfrak p_0)_{\mathfrak p_1}$ we reach the situation where $R$ is a local Noetherian domain of dimension $1$. Then $\mathfrak p_0 = (0)$ and $\mathfrak p_1 = \mathfrak m$ is the maximal ideal.
Translating we have to derive a contradiction from the following: we have a nonzero prime $\mathfrak q \subset \mathfrak m[T]$ with $\mathfrak q \not = \mathfrak m[T]$.
Let $K$ be the fraction field of $R$. Let $\mathfrak q_K \subset K[T]$ be the ideal generated by $\mathfrak q$ in $K[T]$. Then $\mathfrak q = \mathfrak q_K \cap R[T]$.
For every $n \geq 0$ let $R[T]_{\leq n}$ be the polynomials of degree $\leq n$. Let $M_n = \mathfrak q \cap R[T]_{\leq n}$ and $Q_n = R[T]_{\leq n}/M_n$ so that we have a short exact sequence
$$
0 \to M_n \to R[T]_{\leq n} \to Q_n \to 0
$$
Now observe that $Q_n$ is a finite $R$-module, is torsion free, and has rank bounded independently of $n$. Namely, over $K$ we know that $\mathfrak q_K$ is generated by a polynomial of degree $d$ and we see that $Q_n \otimes_R K$ has dimension over $K$ at most $d$.
Pick $a \in \mathfrak m$ nonzero. Then (1) $R/aR$ has finite length $c$, (2) for any finite torsion free module $Q$ of rank $r$ the length of $Q/aQ$ is $rc$, and (3) a module $Q$ with length $Q/aQ$ bounded by $rc$ is generated by $\leq rc$ elements. [Hints for elementary proofs: To prove (1) you show for any $b \in \mathfrak m$ some power of $b$ is in $aR$ otherwise $R/aR$ would have a second prime. To prove (2) you choose $R^{\oplus r} \subset Q$ and you use the snake lemma for multiplication by a on the corresponding ses. To prove (3) use Nakayama and that a finite length module is generated by at most its length number of elements.]
Take $n > dc$ where $d$ is the upper bound for the ranks of all $Q_n$ found above. Then we conclude that there exists an element in $M_n$ which is not in $\mathfrak m(R[T]_{\leq n})$ because we have seen above that $Q_n$ can be generated by $\leq dc$ elements. Small standard argument omitted.
This is the desired contradiction because we assumed $\mathfrak q \subset \mathfrak m[T]$. QED
This answer shows that with usual commutative algebra there is a very short proof. Enjoy!
For any commutative ring $R$ with 1, any projective $R$-module of constant finite rank is finitely generated. This is the content of Exercises I.2.13 and I.2.14 of Weibel's $K$-book.
The argument goes as follows (all the relevant hints are in Weibel's book), I hope I did not introduce any tacit finiteness assumptions.
1) In the constant rank one case, we have the dual $\check{P}=\operatorname{Hom}_R(P,R)$ and the evaluation map $\operatorname{ev}:\check{P}\otimes_R P\to R$ whose image $\tau_P=\operatorname{im}\left(\operatorname{ev}:\check{P}\otimes_RP\to R\right)$ is called the trace. By a result of Kaplansky, $P_{\mathfrak{p}}$ is free for any prime ideal $\mathfrak{p}\subseteq R$, and we can choose a generator $x\in P_{\mathfrak{p}}$. Then $\check{x}\otimes x\mapsto 1$ under $\check{P_{\mathfrak{p}}}\otimes_{R_{\mathfrak{p}}}P_{\mathfrak{p}}\to R_{\mathfrak{p}}$. In particular, $\tau_P\not\subseteq\mathfrak{p}$ since otherwise $P\otimes_R R/\mathfrak{p}=0$ contradicts the constant rank one assumption (as pointed out in the comments of Owen Biesel). Since $\tau_P$ is an ideal, it contains $1$ and we can write $1=\sum f_i(x_i)$. For each $\mathfrak{p}$, some $x_i$ will be a generator of $P_{\mathfrak{p}}$, so the $x_i$ are generators for $P$.
2) If $P$ has constant rank $r$, then $\bigwedge^r P$ has constant rank one and by Step 1 has finitely many generators. These are of the form $\sum a_i y_{1i}\wedge\cdots\wedge y_{ri}$ and the (finitely many) $y_{ij}$ appearing generate $P$.
Best Answer
The answer is yes. Let $I = q_1 \cap q_2 \cap \cdots \cap q_k$ be the primary decomposition of $I$. Let $p_i$ be the radical $\sqrt{q_i}$ and let $N_i$ be large enough that $q_i \supseteq p_i^{N_i}$. I claim that we can take $N= \max(N_i)$. We'll abbreviate $\mathbb{C}[x_1, \ldots, x_n]$ to $A$.
Let $f$ be a polynomial as in the statement of the question. In order to show that $f \in I$, it is enough to show that $f \in q_i$ for every $i$. We focus on one $q$ to pay attention to, so we can drop the subscript $i$ and just talk about $p$, $q$ and $N$. Let $W$ be the variety of $p$.
Now, $A$ is regular, so $A_p$ is regular. In other words, if $p$ has codimension $d$, there are transverse smooth hypersurfaces $y_1$, $y_2$, ..., $y_d$ which generate the maximal ideal of $A_p$. So there is some $u \not \in p$ such that $u^{-1} p = \langle y_1, \ldots, y_d \rangle$ in the localization $A[u^{-1}]$.
We claim that $f$, as an element of $A[u^{-1}]$ is in $u^{-1} p^N$. Proof: Write $$f=\sum_{i_1+\cdots+i_d < N} b_{i_1 \ldots i_d} y_1^{i_1} \cdots y_d^{i_d} + r$$ with $r \in u^{-1} p^N$ and with all the $b$'s either $0$ or not in $p$. We want to show the $b$'s are zero. If not, let $(i_1, \ldots, i_d)$ be minimal with the corresponding $b$ nonzero. Since $b$ is not in $p$, it is not zero on $W$. Choose $w$ in $W$ where neither $b$ nor $u$ vanishes.
Since the $y_i$'s are transverse hypersurfaces, we can take $\sum i_j$ derivatives to obtain a polynomial with leading term $\mbox{nonzero stuff} \cdot \prod (i_j)! \cdot b$ at $w$. Since we are in characteristic zero, this doesn't vanish at $w$. (In finite characteristic, the factorials might be zero.) But $w \in W \subseteq V$, so this derivative is supposed to vanish at $w$. This contradiction shows that $f \in u^{-1} p^N$.
Since $q \supseteq p^N$, we have $f \in u^{-1} q$ so $u^k f \in q$ for some $k$. But $q$ is primary and $u \not \in p = \sqrt{q}$. So this shows $f \in q$, as desired.