I am just posting my comment as one answer. No, you cannot conclude that the image of $t$ is a nonzerodivisor modulo the ideal $I$.

Let $\ell$ and $k$ be positive integers $\geq 2$ with $\ell \geq 2k$. Consider the following polynomial, $$ F=\ell(x_1^{\ell+k}+x_2^{\ell+k})-(\ell+k)t^2x_1^\ell x_2^\ell. $$ The partial derivatives are, $$\partial_1 F = \ell(\ell+k)(x_1^{\ell+k-1}-t^2x_1^{\ell-1}x_2^\ell), \ \ \partial_2 F = \ell(\ell+k)(x_2^{\ell+k-1}-t^2x_1^{\ell}x_2^{\ell-1}),$$ $$\partial_t F = -2(\ell+k)tx_1^\ell x_2^\ell.$$ Thus, modulo the ideal $J=\langle \partial_1 F, \partial_2 F \rangle$, we have the following congruences, $$x_1^{\ell+k-1} \equiv t^2x_1^{\ell-1}x_2^{\ell}, \ \ x_1^{2(\ell+k-1)} \equiv t^4 x_1^{2(\ell-1)}x_2^{2\ell}, \ \ x_2^{\ell + k-1} \equiv t^2 x_1^{\ell}x_2^{\ell-1}, \ \ x_2^{2\ell} \equiv t^2x_1^{\ell} x_2^{2\ell-k}.$$ Taken together, this gives $$ x_1^{2\ell+2k-2} \equiv t^6 x_1^{3\ell-2}x_2^{2\ell-k}, \ \text{i.e.,}\ \ x_1^{2\ell+2k-2}(1-t^6x_1^{\ell-2k}x_2^{2\ell-k})\in J.$$ Since the second factor is invertible in $\mathbb{C}\{x_1,x_2,t\}$, this gives that $x_1^{2(\ell+k-1)}\in J$. By symmetry, also $x_2^{2(\ell+k-1)}\in J$. Therefore, also $(\partial_t F)^3$ is in $J$.

However, $x_1^\ell x_2^\ell$ is not congruent to $0$ modulo $I$. Indeed, the quotient of $I$ by the ideal generated by the image of $t$ equals $$\mathbb{C}\{x_1,x_2,t\}/\langle t,x_1^{\ell+k-1},x_2^{\ell+k-1} \rangle.$$ Since $k\geq 2$, also $k-1\geq 1$, so that $x_1^\ell x_2^\ell$ is nonzero in this quotient ring. Since $tx_1^\ell x_2^\ell$ is in $I$, the image of $t$ modulo $I$ is a zerodivisor.