A lot is known about this problem. 

Firt of all, the answer to your first question is **no**. In fact, we can prove the following easy

>**Proposition.** A  curve $C$ of genus $3$ is hyperelliptic if and only if it contains a theta characteristic $L$ such that $h^0(L) \geq 2$.

>*Proof.* On the canonical model of $C$, an effective theta characteristic is cut out by a contact hyperplane. Since $h^0(L) \geq 2$, we must have infinitely many contact hyperplanes. Thus the canonical model of $C$ cannot be a plane quartic, because any plane quartic contains precisely $28$ bitangent lines. It follows that $C$ is hyperelliptic. $\square$

For the general case, let me just state a single result, referring to [1] and the references given therein for a more complete treatment.

> **Theorem.** Denote by $\mathscr{M}^r_g$ the sublocus of $\mathscr{M}_g$ consisting of curves having a theta characteristic $L$ such that $$h^0(L) \geq r+1 \quad  \textrm{and} \quad  h^0(L) \equiv r \, (\textrm{mod } 2).$$
Then $\mathscr{M}_g^1$ (resp. $\mathscr{M}_g^2$) has pure codimension $1$ (resp. $3$) in $\mathscr{M}_{g}$ if $g \geq 3$ (resp. $g \geq 5$) and a generic point of any of its components is a curve which has only one theta-characteristic $L$ with $h^0(L)=2$ (resp. $h^0(L) =3$ if $g \geq 6$). 

> Moreover if $g \geq 3$ such a $L$ is not composed with an involution (resp. if  $g \geq 6$ such a $L$ has no fixed points). 
 

**References.**

[1] Montserrat Teixidor I Bigas: [Half canonical series on algebraic curves][1], *Trans. Amer. Math. Soc.* **302** (1987), 99-115.


  [1]: http://www.ams.org/journals/tran/1987-302-01/S0002-9947-1987-0887499-X/