Yes, the rank of the Jacobian matrix doesn't depend on the set of generators of $I$. The Jacobian matrix at $x$ represents the subspace generated by the differentials at $x$ of all $f\in I$.

Note that the rank of the Jacobian matrix at $x$ is computed in the fiber where $x$ lives, it has nothing to do with the base scheme.

**Some more explanations** We work over a base field $k$. Let $I$ be the ideal defining $X$ in $Y:=\mathbb A^n$. Then we have a canonical exact sequence
$$ I/I^2 \to \Omega_{Y}|_X \to \Omega_X \to 0 $$
where the first map is $\bar{f}\mapsto df\otimes 1$. Tensoring by $k(x)$ we get
$$ I/I^2 \to \Omega_{Y}\otimes k(x) \to \Omega_X\otimes k(x) \to 0.$$
If $g_1,...,g_m$ are a system of generators of $I$, then $dg_1,...., dg_m$ generate the image of $I/I^2$ in $\Omega_{Y}\otimes k(x)\simeq k(x)^n$. Call this image $C$. Let $J_x$ be the Jacobian matrix associated to $g_1,...,g_m$ at $x$ in a system of coordinates of $Y$. Then the columns of $J_x$ correspond to the images of $dg_1,...,dg_m$ in $C\subseteq \Omega_{Y}\otimes k(x)$. Therefore the rank of $J_x$ is just the dimension of $C$ over $k(x)$, and this is independent on the choice of the system of generators $g_1,...g_m$.

By the way, these discussions show that
$$\dim_{k(x)} (\Omega_{X}\otimes k(x))=n - \mathrm{rank} J_x.$$
As the smoothness at $x$ is equivalent to $ \dim_{k(x)} (\Omega_{X}\otimes k(x))= \dim_x X$, we see that it is also equivalent to $\mathrm{rank} J_x=n-\dim_x X$ which is the Jacobian criterion of smoothness.