There is a non-hyperelliptic (fixed-point free) involution of the surface of genus $5,$ with the quotient a surface of genus $3.$ Further, $A_4$ does come up as the automorphism group thereof, [see][1] 

<cite authors="S. Allen Broughton" mrnumber="1090743" cite="_J. Pure Appl. Algebra_ **69** (1991), no. 3, 233--270">_S. Allen Broughton_, MR 1090743 [**Classifying finite group actions on surfaces of low genus**](http://dx.doi.org/10.1016/0022-4049(91)90021-S), _J. Pure Appl. Algebra_ **69** (1991), no. 3, 233--270.</cite>

So, modulo some care, the answer seems to be YES.

**ADDITION** As pointed out by Noam in the comments, the above is not quite satisfying. The simplest reference is one to t[he paper of Kuribayashi and Kimura][2] 

<cite authors="Akikazu Kuribayashi and Hideyuki Kimura" mrnumber="1068416" cite="_J. Algebra_ **134** (1990), no. 1, 80--103">_Akikazu Kuribayashi and Hideyuki Kimura_, MR 1068416 [**Automorphism groups of compact Riemann surfaces of genus five**](http://dx.doi.org/10.1016/0021-8693(90)90212-7), _J. Algebra_ **134** (1990), no. 1, 80--103.</cite>
Unfortunately, that paper's notation is somewhat hard to penetrate. On the other hand, the way they describe the automorphism group (conjugacy class in $GL()$ should make it easy to check that the relevant surface is not hyperelliptic.

  [1]: https://www.evernote.com/shard/s24/sh/443727bb-27cd-41ca-ba96-8aec175385d1/eb99cdf97d58f161c6a856bfba80eb11
  [2]: https://www.evernote.com/l/ABhoritVc3dDkq1BL_piYVCkr7dKhoZcDdw