The following claim and its proof answers OP's question, but only under an additional hypothesis on the critical points of the coordinate functions $\gamma_i$ of the curve $\gamma = (\gamma_1, \dots, \gamma_n)$.
Claim. Let $n \ge 1$ and let $\gamma = (\gamma_1, \dots, \gamma_n): [0, 1] \rightarrow [0, 1]^n$ be a curve of class $C^1$. Assume moreover that $\gamma_i$ has at most countably many critical points for every $i \in \{1, \dots, n\}$. Then there is $j \in \{1, \dots, n\}$ and $c \in [0, 1]$ such that $$\vert \{ t \in [0, 1] \, \vert \, \gamma_j(t) = c \} \vert \ge L(\gamma) /n$$ where $L(\gamma)$ is the arc length of $\gamma$.
I'll be grateful to contributors willing to answer:
Question. Can the extra hypothesis on critical points be removed, or at least weakened?
The proof of the above claim relies on the next two lemmas.
Lemma 1. Let $\gamma = (\gamma_1, \dots, \gamma_n): [0, 1] \rightarrow [0, 1]^n$ be a rectifiable curve. Then $\gamma_i$ is rectifiable for every $i$ and there is $j \in \{1, \dots, n\}$ such that $L(\gamma_j) \ge L(\gamma) / n$.
Proof. It follows easily from the definition of a rectifiable curve and from the inequalities $\vert x_j \vert \le \sqrt{x_1^2 + \cdots + x_n^2} \le \sum_{i = 1}^n \vert x_i \vert$ that $$L(\gamma_j) \le L(\gamma) \le \sum_{i = 1}^n L(\gamma_i)$$ for every $j \in \{1, \dots, n\}$. Thus there is $j \in \{1, \dots, n\}$ such that $L(\gamma_j) \ge L(\gamma)/n$.
Lemma 2. Let $\gamma: [0, 1] \rightarrow [0, 1]$ be a curve of class $C^1$. Assume moreover that the set of critical points of $\gamma$ is at most countable. Then we have $$\deg(\gamma) \ge L(\gamma)$$ where $\deg(\gamma) = \sup_{y \in [0, 1]} \vert \{ \gamma^{-1}(\{y\})\} \vert$.
Proof. Since $\gamma$ is of class $C^1$, it is rectifiable, i.e., its arc length $L = L(\gamma)$ is finite. Thus we can assume, without loss of generality, that $\deg(\gamma) = d < \infty$. Define $s: [0, 1] \rightarrow [0, L]$ by $s(t) = \int_0^t \vert \gamma'(t) \vert dt$. Clearly, the function $s$ is of class $C^1$. Since the set of critical points of $\gamma$ has empty interior, the function $s$ is strictly increasing and is therefore an homeomorphism. Besides, the function $s$ is differentiable at $s(t)$ if and only if $t$ is not a critical point of $\gamma$, i.e., if and only if $\gamma'(t) \neq 0$. Let $\gamma_1 = \gamma \circ s^{-1}$. Then $\gamma_1$ is a continuous curve which is locally an isometry, i.e., for every $t \in [0, 1]$ such that $\gamma'(t) \neq 0$, there is an open interval $I = I_t$ containing $s(t)$ such that $\gamma_1$ restricts to an isometry of $I$. Let $O \subseteq [0, 1]$ be the complement of the critical values of $\gamma$. As $O$ is open, it is the union of at most countably many disjoint intervals with non-empty interior. Let $J$ be one of these intervals and denote by $\mathring{J}$ its interior. We claim that $\gamma_1^{-1}(\mathring{J})$ is the union of at most $d$ disjoint intervals which are all isometric to $\mathring{J}$. It follows from our claim that $[0, L] = \gamma_1^{-1}([0, 1])$ is the disjoint union of an open set of Lebesgue measure at most $d$, that is $\gamma_1^{-1}(O)$, and a countable set. Therefore $L \le d$.
Proof of the claim. Combine Lemmas 1 and 2.