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**.