If $f:X\to S^n$ is continuous and $v_0\in S^n$, can we find $F:X\to O(n+1)$ such that $F(x)v_0=f(x)$ for all $x$? The set $O(n+1)$ of orthogonal matrices acts transitively on $S^n\subseteq\Bbb R^{n+1}$, meaning that for every $x,y\in S^n$ there is a matrix $A\in O(n+1)$ such that $Ax=y$.
Now suppose that $X$ is a topological space and that we have a continuous map $f:X\to S^n$. If we fix $v_0\in S^n$, the transitivity of $O(n+1)$ means that to every $x\in X$ there exists $A_x\in O(n+1)$ such that $A_xv_0=f(x)$.

Question: Can we chose the matrices $A_x$ in such a way that the map
  $$F:X\to O(n+1),\quad x\mapsto A_x$$
  is continuous?

It would be sufficient to show that the evaluation map
$$\Phi:O(n+1)\to S^n,\quad \Phi(A)=Av_0$$
has a continuous right inverse
$$\Psi:S^n\to O(n+1).$$
In that case we can take $F=\Psi\circ f$.
 A: Your question is actually equivalent to the existence of a right inverse, by taking $X = S^n$. 
Not necessarily. Pick $f: S^2 \to S^2$ to be the identity map, and suppose you had a lift $F: S^2 \to O(3)$. Because $\pi_2 O(3) = 0$, this lift is null-homotopic. Then projecting the null-homotopy down to $S^2$, we obtain a null-homotopy of the original map; that's nonsense, as the identity map $S^2 \to S^2$ induces an isomorphism on second homology.
For $n = 1$, this is true, because the action you've defined $SO(2) \to S^1$ is a diffeomorphism. 
This is, however, occasionally true. One particular case is $n=3$; we may think of $SO(n+1) \to S^n$ as a principal $SO(n)$-bundle, and a section of this bundle is the same thing as a trivialization of the bundle. $G$-bundles over $S^n$ are classified by homotopy classes of maps from the equator $S^{n-1} \to G$; this construction is known as clutching functions. But in particular $\pi_2 SO(4) = 0$, so that this bundle actually is trivial for $n=3$.
A: The right inverse idea you and Mike mentioned is spot on.
Call a sphere $S^n$ special if for every continuous $f:X\rightarrow S^n$, there is a continuous $F:X\rightarrow O(n+1)$ as you've defined it.
Then we have the following theorem:

$S^n$ is special iff $n = 0,1,3,$ or $7$.

Proof:  As mentioned by Mike, $S^n$ is special iff there is a right inverse to the canonical map $O(n+1)\rightarrow S^n$.  But the projection map $O(n+1)\rightarrow S^n$ actually gives $O(n+1)$ the structure of a principal $O(n)$ bundle over $S^n$.  In the context of bundles, the name "right inverse" usually goes by the name "section".
Luckily, it's not too hard to show that a principal bundle has a section iff it's trivial.
Thus, we've shown $S^n$ is special iff the principal bundle $O(n)\rightarrow O(n+1)\rightarrow S^n$ is actually trivial.  But a manifold has a trivial frame bundle iff it is parallelizable.  According to Adams, this occurs iff $n=0,1,3$, or $7$.$\square$
As far as getting an explicit $\Psi$, for $S^0$, we have $O(1) = S^0$, so that's easy.  Further, $O(2)$ is two disjoint copies of $S^1$, so just map $S^1$ to the component of $O(2)$ containing the identity.  (That is, use $SO(2)\subseteq O(2)$ as Mike suggested).
But here is a method that works on all $4$ cases simultaneously.  Let $\mathbb{K}\in \{\mathbb{R}, \mathbb{C}, \mathbb{H}, \mathbb{Ca}\}$, (corresponding to $n=0,1,3,7$ respectively) where the last two are the quaternions and octonions.
Let $e_i\in \mathbb{K}$ denote the vector whose $i$th entry is $1$ with all other entries $0$.
Start by defining $\Psi:S^n\rightarrow O(n+1)$ with $\Psi(e_1)$ to be the matrix in $O(n+1)$ with columns given by $e_1,e_2,..,e_n$, which is just a complicated way of saying $\Psi(e_1) = I$.
Now, given $p\in S^n\subseteq \mathbb{K}$, define $\Psi(p)$ to be the matrix in $O(n+1)$ whose columns are $pe_i$. (In fact, the image is is $SO(n+1)$)
