Let $$\Sigma \subset \mathbb{R}^3$$ be a compact embedded surface with boundary $$\partial \Sigma$$ and $$i:\Sigma\setminus \partial\Sigma \to \mathbb{R}^3 \setminus \partial\Sigma$$ the inclusion.

Is the following true?

If $$i_*(\pi_1(\Sigma\setminus \partial \Sigma))=0$$, then $$\Sigma$$ is orientable.

Yes, the surface is orientable. To simplify the LaTex and the exposition, I will change the notation and setting a small amount.

Suppose that $$F$$ is a compact connected embedded surface in three-sphere. Suppose that the image of $$\pi_1(F)$$ in $$\pi_1(S^3 - \partial F)$$ is trivial. We must show that $$F$$ is orientable.

In the base case, where $$F$$ has no boundary, this follows from Alexander's theorem.

Suppose that $$\alpha$$ is a boundary component of $$F$$. Let $$\alpha'$$ be a curve embedded in $$F$$ which is disjoint from, but is isotopic to (in $$F$$), the boundary component $$\alpha$$. Thus $$\alpha$$ and $$\alpha'$$ cobound an annulus in $$F$$. Since the image of $$\pi_1(F)$$ is trivial in $$\pi_1(S^3 - \partial F)$$, we have that $$\alpha'$$ bounds an immersed disk in $$S^3 - \partial F$$. By Dehn's lemma (that is, by a version of the Disk Theorem) $$\alpha'$$ bounds an embedded disk $$D$$ in $$S^3 - \partial F$$. Thus $$\alpha$$ is an unknot.

This holds for all boundary components of $$F$$. In fact, since all of the boundary components bound disks in $$S^3 - \partial F$$, we deduce that $$\partial F$$ is a split link. Surgering along separating spheres reduces us to the case where $$F$$ has only one boundary component. [There is some work here.]

Now consider the annulus $$A$$ between $$\alpha$$ and $$\alpha'$$. Note that $$A$$ is two-sided. So we may and do isotope $$D$$ slightly to make it transverse to $$F$$ and disjoint from the interior of $$A$$. Suppose that $$\beta$$ (perhaps equal to $$\alpha'$$) is an innermost curve of $$F \cap D$$. Let $$D' \subset D$$ be the subdisk bounded by $$\beta$$. Let $$B \subset F$$ be a small annulus neighbourhood of $$\beta$$.

We surger $$F$$ along $$D'$$ to obtain a new surface $$F'$$. That is, we form $$F - B$$ and glue on a pair of disks, both parallel to $$D'$$. If $$F'$$ is non-orientable, then it still has trivial $$\pi_1$$-image and has lower complexity (either has no boundary, has lower genus, or meets $$D$$ in a simpler way) than $$F$$. This is a contradiction.

We deduce that $$F'$$ is orientable and so is two-sided. We also deduce that the two boundaries of $$B$$ are attached to opposite sides of $$F'$$. Thus there is a (orientation reversing) curve $$\gamma$$ in $$F$$ that meets $$\beta$$ exactly once. We deduce that $$\gamma$$ has linking number one with $$\alpha$$. Thus $$\gamma$$ is non-trivial in the image of $$\pi_1(F)$$, a contradiction.

I think that the condition can be reduced to "$$H_1$$-image is trivial". The above argument does not immediately work (because surgery along an orientable surface can cause genus to increase).