Depending possibly on exactly how you define everything I think the answer is yes in a more general situation:

$\textbf{Proposition:}$ Let $\overline{X}$ be a finite type scheme over $\mathbb{C}$, and $X \subset \overline{X}$ a dense open, and $x \in \overline{X} \setminus X$ a closed point. Then there exists a map $\mathrm{Spec}(\mathbb{C}[[t]]) \to \overline{X}$ such that $(t) \mapsto x$ and $(0) \mapsto y$ for some $y \in X$ which will be the generic point of a curve passing through $x$.

Notice that any map $\mathrm{Spec}(\mathbb{C}[[t]]) \to \overline{X}$ factors through $\mathrm{Spec}(\mathbb{C}[[t]]) \to \mathrm{Spec}(\mathcal{O}_{X,x}) \to \overline{X}$ where $\mathcal{O}_{X,x} \to \mathbb{C}[[t]]$ is a local map (such that $(t) \mapsto x$). Let $A = \mathcal{O}_{X,x}$.

There is a nonempty principal open $D(f) \subset X \cap \mathrm{Spec}(\mathcal{O}_{X,x})$ for some $f \in A$ with $f \notin \mathrm{nilrad}(A)$ but $x \notin D(f)$ since $x \in \overline{X} \setminus X$. Now we apply the following lemma:

$\textbf{Lemma:}$ Let $(A, \mathfrak{m})$ be a Noetherian local ring with finite dimension $\mathrm{dim}(A) \ge 1$ and let $f \in \mathfrak{m}$ with $f \notin \mathrm{nilrad}(A)$. Then there exists a prime ideal $\mathfrak{p} \subset A$ such that $f \notin \mathfrak{p}$ and $\mathrm{dim}(A / \mathfrak{p}) = 1$.

Proof: there is a prime $\mathfrak{p}_0 \subset A$ with $f \notin \mathfrak{p}_0$ because $f \notin \mathrm{nilrad}(A)$. Therefore replacing $A$ by $A / \mathfrak{p}_0$ we can assume $A$ is a domain. Now we proceed by induction on $\mathrm{dim}(A) = n$. If $\mathrm{dim}(A) = 1$ then we have prime ideals $(0) \subsetneq \mathfrak{m}$ so taking $\mathfrak{p} = (0)$ the result follows. Otherwise, choose a maximal length chain of primes,
$$ (0) \subsetneq \mathfrak{p}_1 \subsetneq \mathfrak{p}_2 \subsetneq \cdots \subsetneq \mathfrak{p}_n = \mathfrak{m} $$
Then there are infinitely many prime ideals $\mathfrak{p}_1'$ strictly between $(0)$ and $\mathfrak{p}_2$ which all must have height one since this chain is maximal length (Kaplansky, Commutative Rings, Thm. 144). And thus some $\mathfrak{p}_1'$ does not contain $f$ (Kaplansky, Commutative Rings). Therefore, $A' = A / \mathfrak{p}_1'$ satisfies the hypotheses and $\mathrm{dim}(A') = n - 1$ so we conclude.

Returning to the propositon, we get a prime $\mathfrak{p} \in D(f) \subset X$ with $B = A / \mathfrak{p}$ a Noetherian local domain of dimension one. Let $C$ be the integral closure of $B$ in $\mathrm{Frac}(B)$ so $B \subset C$ is an integral extension of Noetherian domains and thus $\mathrm{dim}(C) = 1$ and $\mathrm{Spec}(C) \to \mathrm{Spec}(B)$ is surjective so there is a maximal ideal $\mathfrak{m}' \subset C$ above $\mathfrak{m}$ and thus $B \to C_{\mathfrak{m}'}$ is a local inclusion of Noertherian domains where $ C_{\mathfrak{m}'}$ is a DVR since it is integrally closed. Then $\widehat{C_{\mathfrak{m}'}} \cong \mathbb{C}[[t]]$ by the Cohen structure theorem. So we get a local injection $B \to C \to C_{\mathfrak{m}'} \to \mathbb{C}[[t]]$. Thus the map,
$$ \mathrm{Spec}(\mathbb{C}[[t]]) \to \mathrm{Spec}(\mathcal{O}_{X,x}/\mathfrak{p}) \to \mathrm{Spec}(\mathcal{O}_{X,x}) \to \overline{X} $$
sends $(t) \mapsto x$ and $(0) \mapsto \mathfrak{p} \in D(f)$.