General position for map from surface to 3-manifold Let f be a smooth map from a (compact,oriented) surface S to a (compact, oriented) 3-manifold M. Suppose that I have an embedded (non-contractible) loop $\gamma$ in my surface $S$, can I find an (immersed) loop $\gamma'$ freely homotopic to $f \circ \gamma$ which is disjoint from $im(S)$? 
 A: In general, no you cannot. Consider one dimension down. Take two curves on a torus, intersecting transversely in a point. One of the curves cannot be homotoped to be disjoint from the pair of curves. Now, cross with a circle $S^1$ to get $T^3$. The pair of curves crossed with $S^1$ is an immersed surface (two immersed tori $T^2$). Then either curve lies in the surface, but may not be homotoped to be disjoint from the immersed surface (this may be proved using the intersection product on homology, dual to the cup product). 
You might object that the surface is disconnected. To obtain a connected surface, just tube them together to get a homologous surface.  
A: Edit: This answer the case of embedded surface, which is different from the question above! 
(Yes. Note that the normal bundle of $S$ in $M$ is an oriented 1-dimensional bundle, in a natural way (by "dividing" the orientation from $M$ by the orientation from $S$). This implies that its restriction to $\gamma$ is trivial, so you can choose a non-zero section $X$. Now you can define the isotopy 
$\gamma_s(t) = Exp_{\gamma(t)}(sX(t))$ for some choice of Riemannian metric say, 
and for $s$ in a small enough neighborhood of $0$ this gives you a disjoint isotopic curve.)  

And now for the general case. In general I think that the answer is no. 
Consider for example the map $S^1 \times S^1 \to S^2$ of degree 1. It can be chosen smooth, e.g. by the standard $"(\theta,\phi)"$ parametrization of the 2-sphere. Now consider the map $S^1 \times S^1 \to S^1 \times S^2$ which is the product of this map with the identity. Let $\gamma$ denote the first coordinate circle in $S^1 \times S^1$. Then the image of $S^1 \times S^1$ in the homology of $S^1 \times S^2$ is $[\{0\} \times S^2]$ because of the assumption on the degree. But the image of $\gamma$ is is clearly $[S^1 \times \{0\}]$, it is really just a parametrization of a coordinare circle. So they intersect non-trivially by the intersection pairing on homology.   
