Assume $\Sigma_1$ and $\Sigma_2$ are two embedded compact surfaces (say orientable) in an orientable 3manifold $M$. Assume $\Sigma_1$ and $\Sigma_2$ are homotopic in $M$. Then are they isotopic?
No, generally they're not. For example, there's only one homotopy class $S^2 \to \mathbb R^3$ but there's two isotopy classes of embeddings (given via how the embedding orients the compact 3manifold it bounds). edit: I think if your 3manifold is irreducible and if your maps $S^2 \to M$ are not null homotopic then the answer is likely yes. But if your 3manifold is say a connect sum of lens spaces then I suspect it's false but I haven't come up with a nice example yet. As Allen points out in the comments below, a connectsum of lens spaces won't work, at least not when your surface is a sphere. edit2: As Misha Kapovich points out, for irreducible 3manifolds and incompressible surfaces homotopy implies isotopy. This is an old theorem of Waldhausen's. "On Irreducible 3manifolds which are sufficiently large" Ann. of Math (2) 87 (1968) 5688. 


If $\Sigma_1 \hookrightarrow M$ is an embedded $\pi_1$injective surface, then any homotopic embedded surface will be isotopic to $\Sigma$. As Ryan and Allen point out, this is due to Waldhausen for incompressible surfaces of genus $>0$, and to Laudenbach for 2spheres, together with the Poincare conjecture. If $\Sigma_1$ is not incompressible in $M$, then there exists $\Sigma_2$ which is homotopic to $\Sigma_1$ but not isotopic. The point is that one may compress $\Sigma_1$ to get a surface $\Sigma'\hookrightarrow M$ which has smaller genus, and then reembed the 1handle (in the same homotopy class) in a knotted fashion to get a nonisotopic surface $\Sigma_2$. Misha observed this in the comments on Ryans question for tori in $S^3$, but it holds more generally. There's an intermediate case of $\Sigma_1\hookrightarrow M$ which is incompressible and not $\pi_1$injective. By the loop theorem, this can only occur if $\Sigma_1$ is 1sided in $M$, which implies that the surface is nonorientable, so does not fall under the purview of your question. I'm not sure if homotopy implies isotopy in this case  I suspect there are 1sided Heegaard surfaces which are homotopic but not isotopic, but I don't know examples off the top. 

