Suppose that $\phi:\mathbb{R}^3 \to S^2$ is of class $\mathscr{C}^1(\mathbb{R}^3\setminus \left\{a\right\}) \cap \mathscr{C}^0(\mathbb{R}^3\setminus \left\{a\right\})$, that is $\phi$ might have a singularity at the point $a \in \mathbb{R}^3$. Suppose that the integral $\int_{\mathbb{R}^3}{e(f)[x]\mathrm{d}x}$ is finite, where $e(f)[x]:=\frac{1}{2}\delta^{\alpha\beta}g_{ij}(\phi(x))\frac{\partial \phi^i}{\partial x_{\alpha}}\frac{\partial \phi^{j}}{\partial x_{\beta}}$ is the energy density of $\phi$ ($\delta^{\alpha\beta}$ is the Kronecker symbol and ($g_{ij}$) is the metric on the three dimensional unit sphere $S^2$).
It is possible to define the degree of $\phi$ around the point $a$ as the degree of the map $\phi$ restricted to a small sphere centered at $a$. Now, I would like to prove that if the energy of $\phi$ (that is the previous integral) is finite, then the degree is $0$.
I was able to prove that only when $|a|$ is sufficiently large. For, since $\phi$ has finite energy, it must approach a constant point on the sphere as $|x|$ approaches infinity, hence the map $\phi$ restricted to a small sphere around $a$ can not be surjective, so the degree is $0$.
Could anybody give me some hints for the general case? Thank you in advance!
You have $\phi$ and $f$ in your question, which appear to mean the same thing.
The value of $a$ is irrelevant, since the map $x\mapsto \phi(x-a)$ has the same energy as $\phi$. You can just consider $a=0$. For this reason, the statement I was able to prove that only when $|a|$ is sufficiently large looks strange. I think what you really meant is that you can prove the result when $r$ is sufficiently large, where $r$ is the radius of the sphere $S_r=\{x:|x-a|=r\}$ to which $\phi$ is restricted.
In that case, you are done, because the degree is independent of $r$. This is the homotopy invariance of topological degree.
Anyway, here is a proof. Suppose, to the contrary, that the degree is nonzero for some $r$. Then it is nonzero for all $r$, by the homotopy invariance. Therefore, $\phi(S_r)=S^2$ for all $r$. By the change of variables formula, the integral of the Jacobian of $\phi_{S_r}$ is at least $4\pi$. Since the energy dominates the 2D Jacobian, the integral of $e(\phi)$ over $S_r$ is bounded from below by a constant $c>0$. Integrating over $r$ from $0$ to $\infty$, we get a divergent integral.