Yes, the counts for $q^n$ ($n \leq g$), together with the value of $q$, are enough.
Let the eigenvalues of Frobenius be $\lambda_i$ ($1 \leq i \leq 2g$), and let $P(t) = \prod_{i=1}^{2g} (1-\lambda t)$ which is a scaling of the characteristic polynomial of Frobenius. Then the power sums $\sigma_n := \sum_{i=1}^{2g} \lambda_i^n$ are the Taylor coefficients of $$ F(t) := -t \frac{P'(t)}{P(t)} = \sum_{i=1}^{2j} \frac{\lambda_i t}{1-\lambda_i t} = \sum_{i=1}^{2j} \left((\lambda_i t) + (\lambda_i t)^2 + (\lambda_i t)^3 + \cdots \right) = \sum_{n=1}^\infty \sigma_n t^n. $$ We are given $\left| A({\bf F}_{q^n}) \right|$, and thus also $\sigma_n = q^n + 1 - \left| A({\bf F}_{q^n}) \right|$, for $n \leq g$.Thus we know $F(t)$ up to $O(t^{g+1})$. Thus we know $-F(t)/t$ up to $O(t^g)$. But that's the logarithmic derivative of $P(t)$, so we know the power series for $\log P(t)$ up to $O(t^{g+1})$ (the constant term vanishes because $P(0)=1$). Thus $$ P(t) = \exp \left(-\!\int_{\tau=0}^t F(\tau) \frac{d\tau}{\tau}\right) $$ gives $P(t)$ up to $O(t^{g+1})$, i.e.\ up to and including the $t^g$ coefficient. Now the functional equation $P(t) = (qt^2)^g P(1/qt)$ gives us the rest: for each $j=1,\ldots,g$, the $t^{g+j}$ coefficient is $q^j$ times the $t^{g-j}$ coefficient. So we know the full expansion of $P$, and thus the full list of eigenvalues and the zeta function, QED.