Here's a half-baked idea involving minimal surfaces (when I have time I will see if I can make this a little more meaningful).

Consider the compact three manifold $M=\Sigma \times \mathbb{S}^1$ with the product metric metric $g+d\theta^2$. This metric has positive scalar curvature. Let $\Pi: M\to \Sigma$ be the natural projection map and observe that the scalar curvature of $M$ satisfies
$$
R_M(p)=R_{\Sigma}(\Pi(p))=2K_{\Sigma}(\Pi(p)).
$$

Obviously, for any $\epsilon>0$ we can perturb the curve $\gamma\subset \Sigma\times \{ 0\}$ to $\gamma_\epsilon$ an embedded curve in $M$ so that $\gamma_\epsilon$ has $C^3$ norm within $\epsilon$ of $\gamma$ and $\Pi(\gamma_\epsilon)=\gamma$. In particular, the geodesic curvature of $\gamma_{\epsilon}$ satisfies $\kappa_{\gamma_\epsilon}=o(\epsilon)$.

Now let $\Gamma_{\mathbb{Z}}(\epsilon)$ be the GMT solution to the Plateau problem with boundary $\gamma_\epsilon$ obtained by minimizing in the space of integer coefficient integral currents and let $\Gamma_{\mathbb{Z}_2}(\epsilon)$ be the solution obtained by minimizing in the space of $\mathbb{Z}_2$ coefficient integral currents. By standard regularity theory, by these are smooth surfaces with boundary $\gamma_\epsilon$ -- $\Gamma_{\mathbb{Z}}$ is orientable while $\Gamma_{\mathbb{Z}_2}$ need not be. Moreover, by standard compactness arguemnts one has that as $\epsilon\to 0$, $\Gamma_{\mathbb{Z}}(\epsilon)\to \Gamma_{\mathbb{Z}}$ where $\Gamma_{\mathbb{Z}}$ is the minimizer (in space of $\mathbb{Z}$-currents) with boundary $\gamma$ and can be thought of as a continuous map from $\Sigma\backslash \gamma\to \mathbb{Z}$ (the function counts multiplicity). Similarly, $\Gamma_{\mathbb{Z}_2}(\epsilon)\to \Gamma_{\mathbb{Z}_2}$ where $\Gamma_{\mathbb{Z}_2}$ is the minimizer in $\mathbb{Z}_2$ currents

Let $p_1, \ldots, p_N$ be the self-intersection points of $\gamma$. By standard boundary regularity estimates, away from $p_1, \ldots, p_N$ the convergence of $\Gamma_{\mathbb{Z}}(\epsilon)$ is smooth (so $|A_{\Gamma_{\mathbb{Z}}(\epsilon)}|=o(\epsilon)$ while by a blow up argument one should have
$$
\int_{\Gamma_{\mathbb{Z}}(\epsilon)\cap B_{\delta}(p_i)} \frac{1}{2}|A_{\Gamma_{\mathbb{Z}}(\epsilon)}|^2 = 2\pi \theta_i +o(\epsilon)+o(\delta)
$$
where here $\theta_i\in [0, 1]$ is related to the angle that $\gamma$ makes at $p_i$ and also to what $\Gamma_{\mathbb{Z}}$ looks like. The same holds for $\Gamma_{\mathbb{Z}_2}(\epsilon)$ though the values of $\theta_i$ may be different.

By the Gauss equations one has
$$
Ric_M(\nu,\nu)=\frac{1}{2} (R_M-K_{\Gamma_{\mathbb{Z}}(\epsilon)}-|A_{\Gamma_{\mathbb{Z}}(\epsilon)}|^2).
$$
Notice $Ric_M(\nu,\nu)=o(\epsilon)$ away from $p_1,\ldots, p_N$ and $Ric_M(\nu,\nu)=O(1)$ every where and one has uniform area bounds.
Hence, integrating this formula, using Gauss Bonnet and the convergence properties above gives
$$
2\pi \geq 2\pi \chi(\Gamma_{\mathbb{Z}}(\epsilon)) = \int_{|\Gamma_{\mathbb{Z}} |} K_{M}-2\pi \sum_{i=1}^N \theta_i +o(\epsilon).
$$
(where the integral over $|\Gamma_\mathbb{Z}|$ means to count multiplicity but not sign.
Similarly,
$$
2\pi\geq 2\pi \chi(\Gamma_{\mathbb{Z}_2}(\epsilon)) = \int_{|\Gamma_{\mathbb{Z}_2} |} K_{M}-2\pi \sum_{i=1}^N \theta_i +o(\epsilon).
$$

I suspect that solving the appropriate Plateau problem for $\gamma$ (I feel $\mathbb{Z}_2$ coefficients might be more enlightening) should give obstructions.

EDIT:

I believe I can use this computation (with $\mathbb{Z}$ coefficients) to rule out the second figure. I may have made a mistake and it is also possible that if this is correct it can still be proved using more elementary arguments...

Label the components of $\Sigma\backslash \gamma$ as follows: To top (i.e. the inverted) lobe is $A$, the left lobe is $B$ the right lobe is $C$ the remaining bounded component (in the figure) is $D$ and the unbounded component is $E$. Denote the self-intersection points $p_A$, $p_B$ and $p_C$ (so $p_A$ is the top self-intersection point, $p_B$ the left one and $p_C$ the right one). Label the angle inside the lobe $A$ at $p_A$, $\phi_A$ and let $\phi_B$ and $\phi_C$ be similar.

If I worked things out correctly, a $\mathbb{Z}$ minimizer must be either $-2A-D+B+C$ or $2B+2C-A+E$.

Lets consider the first case. Blow up analysis implies that the $\theta_A$ is equal to $0$ at $p_A$ in this case (since there are two sheets). On the other hand, $\theta_B=\frac{1}{\pi} \phi_B$ and $\theta_C=\frac{1}{\pi} \phi_C$ Hence, the formula gives
$$
2\pi-4\pi g= 2\int_A K_\Sigma+ \int_B K_\Sigma+\int_C K_\Sigma+\int_D K_\Sigma -2 \phi_B-2\phi_C
$$
Now by Gauss-Bonnet we have
$$
2\pi =\int_{E} K_\Sigma +\pi-\phi_A+ 2\phi_B+2\phi_C
$$
so plugging into the first formula gives
$$
2\pi -4\pi g= 2\int_A K_\Sigma+ \int_B K_\Sigma+\int_C K_\Sigma+\int_D K_\Sigma+\int_{E} K_\Sigma +\pi-\phi_A-2\pi\\=2\pi+\int_A K_\Sigma+\pi-\phi_A
$$
The positive curvature means the RHS is $>2\pi$ and gives a contradiction.

In the second case, one has $\theta_B=\theta_C=0$ and $\theta_A=\frac{1}{\pi} \phi_A$. The formula gives
$$
2\pi-4\pi g= \int_A K_\Sigma+ 2\int_B K_\Sigma+ 2\int_C K_\Sigma +\int_E K_\Sigma -2 \phi_A
$$

Gauss-Bonnet applied to $A\cup D$ gives
$$
2\pi=\int_A K_\Sigma+\int_D K_\Sigma+ \pi-\phi_B+\pi-\phi_C+ \phi_A-\pi
$$
Adding the two equations gives
$$
4\pi(1-g)= 4\pi +\int_{A} K_\Sigma+ \int_B K_\Sigma +\int_C K_\Sigma +\pi-\phi_B+\pi-\phi_C-\phi_A-\pi\\
=10\pi
$$
where the second equality used Gauss-Bonnet to see that
$$
2\pi=\int_A K_\Sigma+ \pi-\phi_A=\int_B K_\Sigma+ \pi-\phi_B=\int_C K_\Sigma+ \pi-\phi_C.
$$
This is also obviously impossible (I'm a little worried I because I didn't use positive curvature in this case).