Yes, they are uniquely determined by $I$. This holds more generally:

Let $R = \oplus_{i \ge 0}R_i$ be a graded ring with $R_0 = K$ a skew field and let $M= \oplus_{i\ge 0}M_i$ be a finitely generated, graded $R$-module. Set $N_k := \sum_{i> 0} R_iM_{k-i}$. If $E$ is a minimal homogeneous set of generators of $M$ over $R$, then, for each integer $k$, the number of elements in $E$ having degree $k$, equals $\dim_K M_k/N_k$.

In particular, this number is independent of $E$. So, if $E,E'$ are two minimal generating sets, they have the same number of elements in each degree (giving a degree-wise bijection) and hence they have the same number of elements altogether.

Proof of the statement: Let the generators be $x_i$ of degree $d_i$. We want to show that $\bar{E}_k := \lbrace x_iN_k \mid d_i = k \rbrace$ is a $K$-basis of $M_k/N_k$.

First note that $M_k = \sum_i R_{k-d_i}x_i= \sum_{d_i \le k}R_{k-d_i}x_i$. Since $\sum_{d_i < k}R_{k-d_i}x_i \subseteq N_k$ and $R_0=K$ it follows that $\bar{E}_k$ is a generating set.

In order to show linear independence, let $a_p \in K$ such that $\sum_p a_px_p = n \in N_k$ (sum over $p$ with $d_p = k)$. By definition $n = \sum_{i+j=k}r_im_j$ with $r_i \in R_i\;(i>0)$ and $m_j \in M_j$. As above we may write $m_j = \sum_l s_{j,l}x_l$ with $\deg(s_{j,l}) = j-d_l$. Putting all together we have

$$\sum_p a_px_p = \sum_{i+j=k}r_i\sum_l s_{j,l}x_l= \sum_l \big (\sum_{i+j=k}r_is_{j,l} \big )x_l=: \sum_l t_l x_l $$
where $\deg t_l=k-d_l \ge 1$, since $\deg r_i \ge 1$. In particular, $\deg x_l = d_l < k$ $= \deg x_p$, so $x_p \neq x_l$ for all $p,l$.

Now, if $\alpha_p \neq 0$ for some $p$, we can express $x_p$ as $R$-linear combination of the $x_i, i \neq p$, contradicting the minimality of $E$. Hence $\alpha_p = 0$ for all $p$.