Bijection between classes of maps from $S^n$ to $\tilde{X}$ and $S^n$ to $X$, where $p:\tilde{X} \to X$ is a convering space 
Let $n > 1$ and let $p:\tilde{X} \to X$ be a covering space, where $X$ and $\tilde{X}$ are path-connected and locally path-connected. Is there a bijection between $\mathrm{Classes}(S^n, \tilde{X}) $ and $\mathrm{Classes}(S^n, X)$, where $\mathrm{Classes}(A,B)$ denotes the homotopy classes of the continuous maps from $A$ to $B$. 

I have tried to create this bijection using lifts: (since $S^n$ is simply-connected for $n > 1$, by the lifting criterion, every continuous map $f:S^n \to X$ admits a lift $\tilde{f}:S^n \to \tilde{X}$). Denote by $[f]$ the class of $f$ in $\mathrm{Classes}(S^n, X)$. The function I thought about was $$[f] \mapsto [\tilde{f}], $$ but this is not well-defined, as $f$ can have many lifts.
EDIT: I have added a few more assumptions that I forgot to write at the beginning.
 A: Caution: I assumed that you were talking about pointed maps in my answer below. That is somehow the more natural thing to talk about here. The general case can be reduced to this one as can be seen in the very nice answer of @John hughes.
For $n = 1$ this is not true. Consider the standard covering $p \colon \mathbb{R} \rightarrow S^1$ given by the exponential. Then we have that $[S^1,\mathbb{R}] = \pi_1(\mathbb{R}) = 1$ is a singleton whereas $[S^1,S^1] = \pi_1(S^1) = \mathbb{Z}$ is not.
For $n \geq 2$, the induced map $p_* \colon \pi_n(\tilde{X}) \rightarrow \pi_n(X)$ is an isomorphism. This follows for example from the long exact sequence of homotopy groups of a fibration (if you don't know what a fibration is just think of a covering here (coverings are certains types of fiber bundles which are specific fibrations). So yes, we have such a bijection at least for all $n \geq 2$.
A: @TMO took us 90% of the way there. Suppose you have a balloon, and you grab the southern hemisphere and squeeze it until all the air is pushed into the northern hemisphere, and the southern hemisphere is reduced to just a line segment. More formally, consider the map 
$$
(x, y, z) \mapsto (\alpha(z)x, \alpha(z)y, z)
$$
where 
$$
\alpha: [-1, 1] \to \Bbb R: t \mapsto \begin{cases} t^2 & t > 0 \\
0 & t \le 0 \end{cases} . 
$$
Then the image of the sphere ends up being the union of a line segment (the "string" on the balloon) and something that's topologically still a sphere. 
Now instead of maps of your original sphere, you can consider maps defined on my collapsed balloon. 
Sorry...I'm rambling here. 
Suppose you have maps $F, G$ from the sphere to $X$. Let me refer to the south pole of the sphere as $S$. Lets say that $Q = F(S)$. If only $Q = G(S)$ as well, then $F$ and $G$ would be pointed-maps, and @TMO's comment would show they're homotopic. So I'm going to define a homotopy from $G$ to a map $G'$ where $G'(S) = Q$, and then we'll be done. 
The idea is to gradually alter $G$ so that it's constant on the southern hemisphere, and then factor through a map to a balloon-and-string-like thing, and then alter G by a homotopy where the end of the string goes to $Q$. I'll need a path $\beta$ from $G(S)$ to $Q$, so $\beta(0) = G(S)$ and $\beta(1) = Q = F(S)$; that exists by path-connectedness. I'll use $\beta$ in a little while. First, let'd define homotopy number 1, using polar coordinates. Express each point of the domain $S^n$ as $(\sin(\phi) \mathbf{v}, \cos(\phi))$, where $\mathbf{v}$ is a vector in $\Bbb R^{n} \subset \Bbb R^{n+1}$. So $\phi$ corresponds to "latitude" in the 2-sphere, with $\phi = 0$ being the north pole, and $\phi = \pi$ being the south pole. OK. Let
$$
H_1((\sin(\phi) \mathbf{v}, \cos(\phi)), t) = \begin{cases}
F(\sin((1+t)\phi) \mathbf{v}, \cos((1+t)\phi)))) & (1+t) \phi \le \pi \\
F(S) & (1+t) \phi > \pi \end{cases}
$$
For $t = 0$, you can see that $H_1(P, 0) = F(P)$, so that's good. For $t = 1$, you can see that $H_1(P, 1)$ is constant on the southern hemisphere (i.e. $\phi \ge \pi/2$), and its value on the equatorial sphere is just $F(S)$. And $P \mapsto H_1(P, 1)$ is homotopic to $F$. I'm going to call that map $F'$. 
Now I'm going to perform another homotopy: holding the value on the equatorial sphere to be $F(S)$, I'm going to drag the image of the south pole over to $Q$. Here goes: 
$$
H_2((\sin(\phi) \mathbf{v}, \cos(\phi)), t) = \begin{cases}
F(\sin(2\phi) \mathbf{v}, \cos(2\phi)) & \phi \le \pi/2 \\
\beta(t (\frac{2}{\pi}\phi - 1) & \phi > \pi/2 \end{cases}.
$$
Let's see what's going on in that second line. We have $\pi/2 \le \phi \le \pi$ in the southern hemisphere. Multiplying by $2/\pi$, we get $1 \le \frac{2}{\pi} \phi \le 2$; subtracting $1$ gets us a number that's $0$ at the equator and 1 at the south pole. 
We multiply by $t$, so when $t = 0$, the argument to $\beta$ is $0$, and $\beta(0) = F(S)$, which is great. In fact, for all $t$, when $\phi = \pi/2$, we end up computing $F(S)$. But as $t$ increases to $1$, the argument to $\beta$ gradually increases to something that ranges from $0$ to $1$ as $\phi$ varies from $\pi/2$ to $\pi$. And $\beta(1)$ is exactly $Q$. 
So $H_2$ provides a homotopy from $F'$ to a new function $F''$ for which we have $F''(S) = Q$. And now @TMO's argument shows that $F''$ is homotopic-rel-basepoints to $G$, so we're done!
(Man, that's hard without a whiteboard!)
A: $p_* : [\mathbb{S}^n, \tilde{X}] \to [\mathbb{S}^n, X]$ does not have to be injective for $n \geq 2$ if you forget the basepoint.
For example for $n=2$, take $X$ the union of a sphere $\mathbb{S}^2$ and a diameter. Then for $\tilde{X}$ take the universal cover of $X$, which is a sequence of infinitely many spheres $\mathbb{S}^2$, where a sphere is connected to the next sphere with a line segment.
Then take $f,g : \mathbb{S}^2 \to \tilde{X}$ to be two maps that map to different spheres in the universal cover, then these two maps are not homotopic. However $p \circ f = p \circ g$, so the images are homotopic.
