$\def\ZZ{\mathbb{Z}}\def\FF{\mathbb{F}}$This is an attempt to fix a (currently deleted) answer of Will Sawin's. Will tried to show that the claim was false for $n = 3^g$. I can show, by a similar strategy, that it is false for $n=3^{2g+\binom{2g}{2}-1}$ and $g \geq 3$.
Here is the motivating idea. Let $G_1 = \pi_1(\Sigma)$ and let $G_1 \triangleright G_2 \triangleright G_3 \triangleright \cdots$ be the lower central series; see here for an explicit. If $\gamma$ is a non-separating loop in $\Sigma$, then the image of $[\gamma]$ is a nontrivial, primitive, vector in $G_2/G_2$. If $[\gamma]$ is a separating loop, then $[\gamma]$ is in $G_2$ and the image of $\gamma$ is a nontrivial primitive vector in $G_2/G_3$. Roughly speaking, we will construct two representations of $\pi_1(\Sigma)$ which have the same characteristic polynomial at every element not in $G_3$.
The details will be messier because I need to work with $G_1/G_2$ and $G_2/G_3$ "modulo $p$", and I don't know enough group theory to say this the right way, so I worked out a bunch by hand the wrong way.
Let $p$ be an odd prime (eventually, $p$ will be $3$). Let $V = H_1(\Sigma, \FF_p)$. Let $\Lambda$ be the exterior algebra $\bigoplus_{j=0}^{2g} \bigwedge^j V$ and let $\Lambda_{\leq 3}$ be the quotient of $\Lambda$ by the two-sided ideal $\bigoplus_{j=4}^{2g} \bigwedge^j V$. Let $\omega$ be the element $\sum_{i=1}^g a_i \wedge b_i$ in $\bigwedge^2 V$, where $a_1$, $a_2$,
..., $a_g$, $b_1$, $b_2$, ..., $b_g$ is a standard basis of $V$ and let $R$ be the quotient of $\Lambda_{\leq 3}$ by the two-sided ideal generated by $\omega$.
We'll write $R_0$, $R_1$, $R_2$, $R_3$ for the graded pieces of $R$, and $R_{\geq j}$ for $R_j \oplus R_{j+1} \oplus \cdots \oplus R_3$.
Our next task is to construct a homomorphism $\phi$ from $\pi_1(\Sigma)$ to the unit group of $R$.
Choose a presentation of $\pi_1(\Sigma)$ with generators $A_1$, $A_2$, ..., $A_g$, $B_1$, $B_2$, ..., $B_g$ and relation
$$\prod_{i=1}^g (A_i B_i A_i^{-1} B_i^{-1}) = 1.$$
Define $a_i$ and $b_i \in V$ to be the images of $A_i$ and $B_i$ under the identification $V = \pi(\Sigma)^{\text{ab}} \otimes \FF_p$.
Lemma 1 The map $A_i \mapsto 1+a_i$, $B_i \mapsto 1+b_i$ extends to a homomorphism $\phi: \pi(\Sigma) \to R^{\times}$.
Proof For any $u \in V$, we have $(1+u)^{-1} = 1-u$ in $\Lambda$. For any $u$ and $v \in V$, we have
$$(1+u) (1+v) (1+u)^{-1} (1+v)^{-1} = 1+2 u \wedge v$$
in $\Lambda$. So
$$\prod_{i=1}^g ((1+a_i) (1+b_i) (1+a_i)^{-1} (1+b_i)^{-1}) = \prod_{i=1}^g (1+2 u_i \wedge v_i)$$
in $\Lambda$. In $\Lambda_{\geq 3}$, we can expand this as $1+2 \sum a_i \wedge b_i$, so in $R$ it is $1$. $\square$
Note that, although the specific map $\phi: \pi_1(\Sigma) \to R^{\times}$ depends on a choice of presentation of $\pi_1(\Sigma)$, we have the following identities for any elements of $\pi_1(\Sigma)$:
Lemma 2 If $U$ is in $\pi_1(\Sigma)$, and $u$ is its image in $V$, then $\phi(U) \equiv 1+U \bmod R_{\geq 2}$.
Lemma 3 If $X_1$, $X_2$, ..., $X_k$, $Y_1$, $Y_2$, ..., $Y_k$ are in $\pi_1(\Sigma)$, and $x_1$, $x_2$, ..., $x_k$, $y_1$, $y_2$, ..., $y_k$ are their images in $V$, then
$$\phi\left( \prod (X_i Y_i X_i^{-1} Y_i^{-1}) \right) = 1 + 2 \sum x_i \wedge y_i \bmod R_{\geq 3}.$$
We thus deduce
Lemma 4: Let $C$ be a nonseparating simple curve in $\Sigma$. Then $\phi(C)$ is nontrivial modulo $R_{\geq 2}$.
Proof: Let $c$ be the corresponding element of $V$. Since the class of $C$ in $H_1(\Sigma)$ is primitive, we know that $c$ is nonzero. Then $C \equiv 1+c \bmod R_{\geq 2}$ by Lemma 2. $\square$
Lemma 5: Let $C$ be a nonseparating simple curve in $\Sigma$. Then $\phi(C)$ is nontrivial modulo $R_{\geq 3}$.
Proof: Let $C$ separate $\Sigma$ into a curve of genus $k$ with one puncture, and a curve of genus $g-k$ with one puncture. Then
$$C = \prod_{i=1}^k (X_i Y_i X_i^{-1} Y_i^{-1}) = \left( \prod_{j=k+1}^g (X_j Y_j X_j^{-1} Y_j^{-1}) \right)$$
for some standard generators $X_i$, $Y_i$ of $\Sigma$. Let $x_i$ and $y_i$ be the corresponding elements of $V$. Then
$$\phi(C) \equiv 1 + 2 \sum_{i=1}^k x_i \wedge y_i \equiv 1 - 2 \sum_{j=k+1}^g x_j \wedge y_j. \bmod R_{\geq 3}$$
Since $R_2$ is $\bigwedge^2 V {\big /} \FF_p \omega$, and $\omega$ has rank $g$, we know that $\sum_{i=1}^k x_i \wedge y_i$ is not a multiple of $\omega$. $\square$
So, at this point we know that any simple curve maps to a class which is not $1 \bmod R_{\geq 3}$. It remains to construct two representations of $\pi_1(\Sigma)$ which agree on classes which are not $1 \bmod R_{\geq 3}$.
Let $U_j$ be the group $\{ u \in R^{\times} : u \equiv 1 \bmod R_{\geq j} \}$. Choose two nontrivial characters $\chi_1$ and $\chi_2$ of $U_3$, and let $W_1$ and $W_2$ be their inductions to $U_1$. Consider the $W_i$ as representations of $\pi_1(\Sigma)$ by pulling back along $\phi$. For $g \in U_1$ and $\not\in U_3$, we have $g^p = 1$ (even when $p=3$) and the character of $W_i$ on $g$ is $0$, since none of the conjugates of $g$ are in $U_3$. However, if $g \in U_3$, then the character of $W_i$ on $g$ is $[U_1:U_3] \chi_i(g)$ so, if $\chi_1(g) \neq \chi_2(g)$, then the characters of $W_1$ and $W_2$ differ at $g$.
The dimension of $W_i$ is $[U_1:U_3] = \#(R_{\geq 1}/R_{\geq 3}) = p^{2g + \binom{2g}{2} - 1}$, as claimed.
