Does an analogue of the sphere theorem hold in higher dimensions?

An embedded codimension 1 sphere in a manifold with $H^1(M;\Bbb Z/2) = 0$ (so that every codimension 1 manifold splits $M$ into two components) gives a splitting of your manifold as a connected sum, so if your manifold does not have a non-trivial such splitting then all embedded spheres are null-homotopic. By the resolution of the Poincare conjecture, $S^n \times S^2$ gives a counterexample in dimension $n+2$ for all $n \geq 2$.

I can't seem to add details on the other answer so I will do so here.

First, it is not necessary to know that $\pi_{n+1} S^2$ is nonzero --- you just need to know that $\pi_{n+1} S^n$ is, which is either $\Bbb Z$ for $n=2$ or $\Bbb Z/2$ for $n>2$.

Second, let me phrase the cohomology of manifolds in a way that makes it obvious that $S^n \times S^2$ is irreducible. To every compact oriented manifold without boundary is associated a graded group $M^i(X) = H^i(X;\Bbb Z)$ when $0 < i < \dim X$, and zero otherwise; $M$ stands for middle-dimensional cohomology.

Then Poincare duality says that the cup-product pairing is a degree $n$ unimodular perfect pairing $M^*(X) \otimes M^*(X) \to \Bbb Z$. We also know that Mayer-Vietoris shows that $M^*(X \# Y) = M^*(X) \oplus M^*(Y)$, where under this direct sum decomposition the cup product decomposes as $$\smile_{X \# Y} = \smile_X \oplus \smile_Y.$$

So to show that some manifold $X$ is irreducible under connected sum, it will suffice that its intersection form is irreducible under directed sum. But $$M^*(S^2 \times S^n) = \Bbb Z \oplus \Bbb Z$$ with cup-form $$\smile_{S^2 \times S^n} = \begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix}.$$ It is easily seen that this is irreducible under direct sum: It would need to decompose into the sum of two 1-dimensional forms, in which case one of those forms should take the value `1' on some vector. But $(a,b) \smile (a,b) = 2ab$, which is always even --- so no such decomposition.

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No idea about immersions. Sounds hard.