Let f be a smooth map from a (compact,oriented) surface S to a (compact, oriented) 3manifold M. Suppose that I have an embedded (noncontractible) loop $\gamma$ in my surface $S$, can I find an (immersed) loop $\gamma'$ freely homotopic to $f \circ \gamma$ which is disjoint from $im(S)$?
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.
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 1dimensional 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 nonzero 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 2sphere. 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 nontrivially by the intersection pairing on homology.

$\begingroup$ Sorry my question maybe lacked some precision. I only assume $f: S \to M$ is a smooth map and I want to make $\gamma$ disjoint from im(S). $\endgroup$ – algebrachallenged Feb 25 '18 at 21:53

$\begingroup$ oh now I see that it is really what you asked im sorry for the irrelevant answer. $\endgroup$ – S. carmeli Feb 25 '18 at 21:59

$\begingroup$ well, maybe one needs to know whether the 2  homology class in M which is the image of the fundamental class of S is trivial or not mod 2. or rather what the intersection number is in M of the degree 2 class represented by f([S]) and the degree 1 class represented by f[gamma]). If they meet, then you should not be able to move the curve off f(S). just playing here along the same idea as S. carmeli. $\endgroup$ – roy smith Feb 25 '18 at 22:04

$\begingroup$ you might try to deform f first, then use S. carmeli's idea, e.g. you should be ok if f deforms to an embedding. $\endgroup$ – roy smith Feb 25 '18 at 22:50