It does not necessarily follow that the bundle is trivial.
In fact, a counterexamaple can be found in Kamerich's thesis
B. N. P. Kamerich, Transitive transformation groups of products of two spheres, Ph.D. thesis, Catholic University of Nijmegen, 1977.
He proves the following.
Theorem: Let $G = SU(n+1)\times SU(2)$. Let $J = SU(n+1)\times S^1$ and let $H = SU(n)\times S^1_{p,q}$, where the embedding $J\rightarrow G$ is the obvious one and the embedding $H\rightarrow G$ maps $(A,z)\in SU(n)\times S^1_{p,q}$ to the block diagonal $$\left( \operatorname{diag}(z^p A, z^{-np}), \operatorname{diag}(z^q, z^{-q})\right)$$ where $p$ and $q$ are relatively prime integers. Assume $q|n$, $n$ is even, $n/q$ is odd, and that $p$ is odd. Then $G/H$ is the (unique) non-trivial linear $S^{2n+1}$ bundle over $S^2$.
(If you want a specific example, take $n = 6$, $q=2$, and $p = 1$).
In fact, not only is the bundle non-trivial, but Kamerich shows the total space is not even homotopy equivalent to $S^{2n+1}\times S^2$ - the second Stiefel-Whitney class is non-trivial for $G/H$ (but it is trivial for the parallelizable manifold $S^{2n+1}\times S^2$.)
Edit There are still examples where the total space has even dimension, simply by producting the total space and fiber of the above example with your favorite odd dimensional homogeneous space.
For example, given $G,J,H$ as above, consider $G' = G\times SU(2)$, $J' = J\times SU(2)$ (with $J\subseteq G$, $SU(2)\subseteq SU(2)$), and keep the same $H$ (still embedded in $J$ as above).
Then the homogeneous fibration is simply $S^{2n+1}\times SU(2)\rightarrow (G/H)\times SU(2)\rightarrow S^2$. The total space still has a non-trivial second Stiefel-Whitney class, so cannot be homotopy equivalent to $S^{2n+1}\times SU(2)\times S^2$.
However, I claim
Given a homogeneous fibration $J/H\rightarrow G/H\rightarrow G/J$ with $G/J = S^2$ and $J/H$ of non-zero Euler characteristic, then the bundle must be trivial.
Proof: Let's handle the case of $G$ simple first. Then the long exact sequence in homotopy groups associated to $J\rightarrow G\rightarrow S^2$ shows that the map $\pi_3(J)\rightarrow \pi_3(G)\cong \mathbb{Z}$ is trivial. In particular, it follows that $J$ is a torus. Since the higher homotopy groups of $J$ then vanish, the fact that $\pi_k(S^2)$ is torsion for all $k > 4$ implies the same of $G$. However, the rational homotopy groups of Lie groups are known, so this implies $G = SU(2)$ (up to finite cover), which then implies $J = S^1$. But then the only non-trivial $H\subseteq J$ of full rank (so that $J/H$ has non-zero Euler characteristic) is $H = J$. Thus, the fibration is the trivial fibration $\{pt\}\rightarrow S^2\rightarrow S^2$, so is obviously a product.
So we may assume (by passing to a finite cover) $G = G_1 \times G_2 \times ...\times G_m \times T^n$ is a product of semi-simple groups and a torus. Because the fiber has non-zero Euler characteristic, the same is true of $G/H$, so $H$ must have full rank in $G$. A theorem of Borel asserts then that $H$ splits as a product $H = H_1\times H_2\times ...\times H_m\times T^n$ with each $H_i\subseteq G_i$ of maximal rank. In particular, the torus factor of $G$ plays no role (since it's a common normal subgroup to both $H$, $J$, and $G$), so we may as well assume $n = 0$. In fact, if any $H_i = G_i$, then necessarily the action of $G_i$ on the homogeneous space $G/H$ and $G/J$ is trivial, so we may as well exclude it. That is, we may assume that $H_i$ is a proper subgroup of $G_i$ for each $i$.
Likewise, $J$ splits as $J = J_1\times J_2\times...\times J_m$. Now, $G/J = (G_1/J_1)\times (G_2/J_2)\times ... \times (G_m/J_m)$. Since $\pi_1(G/J) = 0$, $\pi_1(G_k/J_k) = 0$ for each $k$. This implies the Euler characteristic of $G_k/J_k$ is at least $2$ whenver $J_k\neq G_k$. Since $\chi(S^2) = 2$, we conclude that $J_k = G_k$ for all $k$ but one. By relabeling, we may as well assume that the "one" is $k = 1$. That is, $J_k = G_k$ for $k = 2,..., m$, and thus $G_1/J_1 = S^2$. By the simple case we did above, this means that $G_1 = SU(2)$ and $J_1 = S^1$.
What about $H$? Well, $H_1\subseteq J_1 = S^1\subseteq G_2 = SU(2)$, and $H_1$ must have full rank. Thus $H_1 = J_1 = S^1$. The other $H_i\subseteq G_i$ are arbitrary, except for having full rank.
It now follows that the homogeneous fibration really looks like $$(S^1/S^1)\times (G_2/H_2)\times ... \times (G_m/H_m)\rightarrow (SU(2)/S^1)\times (G_2/H_2)\times ....\times (G_m/H_m)\rightarrow (SU(2)/S^1)\times (G_2/G_2)\times ...\times (G_m/H_m)$$ with the obvious projection.
Canceling out any common normal subgroups and writing $S^2 = SU(2)/S^1$, $F = (G_2/H_2)\times ...\times (G_m/H_m)$, this is simply $$F\rightarrow S^2\times F\rightarrow S^2$$ with the obvious projection. $\square$