Suppose $J :TM \to TM$ is an almost complex structure on $M$, and let $\varphi : TM \to TM$ be a bundle automorphism. Set $J_{\varphi} = \varphi\circ J\circ \varphi^{-1}$, then
\begin{align*}
J_{\varphi}\circ J_{\varphi} &= \varphi\circ J\circ \varphi^{-1}\circ\varphi\circ J\circ \varphi^{-1}\\
&= \varphi\circ J\circ J\circ \varphi^{-1}\\
&= \varphi\circ (-\operatorname{id}_{TM})\circ\varphi^{-1}\\
&= -\varphi\circ\operatorname{id}_{TM}\circ\varphi^{-1}\\
&= -\operatorname{id}_{TM}
\end{align*}
where we have used the fact that $\varphi$ is linear on fibres in the penultimate equality.
Therefore, if $M$ admits an almost complex structure $J$, every bundle automorphism $\varphi$ gives rise to another almost complex structure $J_{\varphi}$, but not every almost complex structure on $M$ arises in this way; for example, $-J$.
Note, the previous paragraph is nothing more than a global version of the following linear algebra statement: if $J$ is a matrix which squares to $-I_n$, then any matrix which is similar to $J$ also squares to $-I_n$, but not every such matrix is similar to $J$; for example, $-J$.
In Sur les classes caractéristiques des structures fibrés sphériques, Wu proved that $S^{4n}$ cannot admit an almost complex structure for $n \geq 1$.
In Groupes de Lie et puissances réduites de Steenrod, Borel and Serre proved that $S^{2n}$ cannot an admit complex structure for $n \geq 4$.
Therefore, the only even-dimensional spheres which can admit an almost complex structure are $S^2$ and $S^6$. In fact, both of these do admit almost complex structures. The former is the familiar Riemann sphere, while the latter obtains an almost complex structure by viewing it as the unit length purely imaginary octonions. Note, one can obtain the almost complex structure on $S^2$ in an analogous way using quaternions; see this question.
The almost complex structure on $S^2$ is integrable; in fact, every almost complex structure on a two-dimensional manifold is integrable, see this question. However, the almost complex structure on $S^6$ described above is known to be non-integrable. It is still unknown whether $S^6$ admits an integrable almost complex structure. In fact, it is unknown whether there exists a $2n$-dimensional manifold, $n \geq 3$, which admits almost complex structures without admitting an integrable almost complex structure. In dimension $4$, there are examples of manifolds which admit almost complex structures, none of which is integrable; one example is $(S^1\times S^3)\#(S^1\times S^3)\#\mathbb{CP}^2$, see Theorem $9.2$ of Barth, Hulek, Peters, & van de Ven, Compact Complex Surfaces (second edition).
See my answer here for an update on the search for an integrable almost complex structure on $S^6$.
Best Answer
The point is that $V = \mathbb{R}^3 = \operatorname{Im}\mathbb{H}$ and $V = \mathbb{R}^7 = \operatorname{Im}\mathbb{O}$ inherit a cross-product $V \times V \to V$ from quaternion and octonion multiplication. It is given by $u \times v = \operatorname{Im}(uv) = \frac{1}{2}(uv - vu)$.
Given an oriented hypersurface $\Sigma \subset V$ with corresponding Gauß map $\nu \colon \Sigma \to S^n$ ($n \in 2,6$) sending a point $x \in \Sigma$ to the outer unit normal $\nu(x) \perp T_x\Sigma$ one obtains an almost complex structure by setting $$J_x(u) = \nu(x) \times u\qquad\text{for }u \in T_x\Sigma.$$
One economic way to see that this is an almost complex structure: The cross-product is bilinear and antisymmetric. It is related to the standard scalar product via $$ \langle u \times v, w\rangle = \langle u, v \times w\rangle $$ (which shows that $u\times v$ is orthogonal to both $u$ and $v$) and the Graßmann identity $u \times (v \times w) = \langle u,w\rangle v - \langle u,v\rangle w$ (only valid in dimension $3$) has the variant $$ (u \times v) \times w + u \times (v\times w) = 2\langle u,w\rangle v-\langle v,w\rangle u - \langle v,u\rangle w $$ valid in $3$ and $7$ dimensions (which shows that $u \times (u \times v) = -v$ for $u \perp v$ and $|u| = 1$).
Therefore $J_x(u) = \nu(x) \times u$ indeed defines an almost complex structure $J_x \colon T_x\Sigma \to T_x\Sigma$.
Specializing this to $\Sigma = S^2$ embedded as unit sphere in $\operatorname{Im}\mathbb{H}$ you can “see” (or calculate) that $J_x$ acts in exactly the same way as multiplication by $i$ on $\mathbb{CP}^1$.