That integral is zero because it's over all space. As you mentioned, by the divergence theorem:

$$\int_{\mathbb R^3} d^3 \mathbf x \boldsymbol\nabla . \big(\phi(\mathbf x) \boldsymbol \nabla \phi(\mathbf x)\big) = \oint_{\partial V} da \ \phi(\mathbf x) \boldsymbol \nabla \phi(\mathbf x)$$ Now if your integral is over all space, which it is, the boundary is "at infinity". The potential goes to zero at infinity; making the right boundary integral go to zero. Thus, $$\int_V d^3 \mathbf x \boldsymbol\nabla . \big(\phi(\mathbf x) \boldsymbol \nabla \phi(\mathbf x)\big)=0$$

That integral is zero because it's over all space. As you mentioned, by the divergence theorem:

$$\int_{\mathbb R^3} d^3 \mathbf x \boldsymbol\nabla . \big(\phi(\mathbf x) \boldsymbol \nabla \phi(\mathbf x)\big) = \oint_{\partial {\mathbb R^3}} d \mathbf a \ .\phi(\mathbf x) \boldsymbol \nabla \phi(\mathbf x)$$ Now if your integral is over all space, which it is, the boundary is "at infinity". The potential goes to zero at infinity; making the right boundary integral go to zero. Thus, $$\int_{\mathbb R^3} d^3 \mathbf x \boldsymbol\nabla . \big(\phi(\mathbf x) \boldsymbol \nabla \phi(\mathbf x)\big)=0$$