There exists a simply-connected closed $6$-manifold $M$ with a homotopy class $\alpha\in \pi_4(M)$ which does not contain an immersion. The following argument is due to Diarmuid Crowley, after we realized that my argument with Stiefel-Whitney classes could not produce examples.

According to Wall,

Wall, C. T. C., Classification problems in differential topology. V: On certain 6- manifolds, Invent. Math. 1, 355-374 (1966); Corrigendum. Ibid. 2, 306 (1967). ZBL0149.20601

there exists a $1$-connected closed $6$-manifold $M$ such that $H^*(M)\cong H^*(\mathbb{C}P^3)$, the cup product $H^2(M)\times H^2(M)\to H^4(M)$ is trivial, and the first Pontryagin class $p_1(M)\in H^4(M)$ is non-zero. The relevant classification result is Theorem 5 in the linked paper.

Claim: The Hurewicz map $\pi_4(M)\to H_4(M)$ is onto.

Proof: $M$ has the homotopy type of a CW-complex with one cell in each even dimension betoween $0$ and $6$. Since the cup square of the generator in $H^2(M)$ is trivial, it follows that the attaching map $S^3\to S^3$ of the $4$-cell has trivial Hopf invariant, therefore is trivial. So $M^{(4)}\simeq S^2\vee S^4$. The claim follows.

Now let $\alpha\in \pi_4(M)$ be a class whose Hurewicz image is a generator $x\in H_4(M)$. Suppose $\alpha$ is represented by an immersion $f:S^4\looparrowright M$. The normal bundle of $f$ is a rank $2$ bundle $\nu(f)$ over $S^4$, hence is trivial. Also, $TS^4$ is stably trivial. The isomrphism of bundles $\nu(f)\oplus TS^4 \cong f^*(TM)$ and naturality of Potryagin classes now implies that $f^*(p_1(M))=0\in H^4(S^4)$. Thus $$ 0=\langle f^*(p_1(M)),[S^4]\rangle = \langle p_1(M),f_*[S^4]\rangle = \langle p_1(M),x\rangle $$ which implies $p_1(M)=0$, a contradiction.

There exists a simply-connected closed $6$-manifold $M$ with a homotopy class $\alpha\in \pi_4(M)$ which does not contain an immersion. The following argument is due to Diarmuid Crowley, after we realized that my argument with Stiefel-Whitney classes could not produce examples.

According to Wall,

Wall, C. T. C., Classification problems in differential topology. V: On certain 6- manifolds, Invent. Math. 1, 355-374 (1966); Corrigendum. Ibid. 2, 306 (1967). ZBL0149.20601

there exists a $1$-connected closed $6$-manifold $M$ such that $H^*(M)\cong H^*(\mathbb{C}P^3)$, the cup product $H^2(M)\times H^2(M)\to H^4(M)$ is trivial, and the first Pontryagin class $p_1(M)\in H^4(M)$ is non-zero. The relevant classification result is Theorem 5 in the linked paper.

Claim: The Hurewicz map $\pi_4(M)\to H_4(M)$ is onto.

Proof: $M$ has the homotopy type of a CW-complex with one cell in each even dimension betoween $0$ and $6$. Since the cup square of the generator in $H^2(M)$ is trivial, it follows that the attaching map $S^3\to S^2$ of the $4$-cell has trivial Hopf invariant, therefore is trivial. So $M^{(4)}\simeq S^2\vee S^4$. The claim follows.

Now let $\alpha\in \pi_4(M)$ be a class whose Hurewicz image is a generator $x\in H_4(M)$. Suppose $\alpha$ is represented by an immersion $f:S^4\looparrowright M$. The normal bundle of $f$ is a rank $2$ bundle $\nu(f)$ over $S^4$, hence is trivial. Also, $TS^4$ is stably trivial. The isomrphism of bundles $\nu(f)\oplus TS^4 \cong f^*(TM)$ and naturality of Potryagin classes now implies that $f^*(p_1(M))=0\in H^4(S^4)$. Thus $$ 0=\langle f^*(p_1(M)),[S^4]\rangle = \langle p_1(M),f_*[S^4]\rangle = \langle p_1(M),x\rangle $$ which implies $p_1(M)=0$, a contradiction.