Fix a hyperkähler manifold $X$ and an identification of $S^2$ with the hyperkähler sphere of $X$. Now consider the twistor space $T := S^2\times X$ equipped with the tautological complex structure. For each $x\in X$, we have a holomorphic map $u_x:S^2\to T$ defined by $u_x(\theta):=(\theta,x)$.
Question: Is every holomorphic map $u:S^2 \to T$ which satisfies ${\rm pr}_1\circ u={\rm id_{S^2}}$ of this form?
I think the preprint arXiv:1006.0440 of Jardim and Verbitsky will answer your question. In short the answer is no, since if $\dim_C X = n$ then the deformation space of sections has dimension $\dim H^0(S^2,N_{S^2/S^2\times X}) = \dim H^0(P^1,O_{P^1}(1)^n) = 2n$ which is twice the dimension of the space which you consider.
As explained by Sascha, the answer to this question is no. But there is a reformulation of the question as follows: The twistor space has a natural anti-holomorphic involution given by $\rho(I,x)=(-I,x)$ for $(I,x)\in S^2\times X.$ Then you can ask whether all holomorphic sections which are real, i.e. $s(-I)=\rho(s(I)),$ are twistor lines, i.e. of the form $I\mapsto (I,x)$ for some $x\in X,$ see my question my question on MO. The answer to the question is again no, but such examples are more complicated to construct then the deformations described in Sascha's answer.
