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.