Clarifying divergence Theorem knowledge I understand the Divergence theorem is as follows, 
$\iiint\limits_{v} (\nabla \bullet \textbf{F}) \;dV= \iint\limits_{dv} \textbf{F} \bullet d \textbf{S}$
My query is if i already know $(\nabla \bullet \textbf{F})= 0$ then can i assume the following,
$(\nabla \bullet \textbf{F})\iiint\limits_{v} \;dV= \iint\limits_{dv} \textbf{F} \bullet d \textbf{S}= 0$.
 A: No.
Let's look at your equation:
$$
\nabla \cdot \mathbf F \iiint_vdV=\iint_{dv}\mathbf F\cdot dS
$$
Putting the meaning of the divergence theorem aside, let's just see the values of the LHS and RHS in your equation. The LHS is a function because $\nabla\cdot \mathbf F$ is a function of $(x,y,z)$. However, the RHS is a number. Obviously, the two quantities cannot be equal.
A: The Divergence Theorem states that given $V\subseteq\mathbb{R^n}$ which has a smooth boundary $S$, and given a differentiable vector field $\mathbf{F}$ (of class $C^1$) which is continuous in a neighborhood  of $V$, then the following is true:

$$\iiint_V\left(\nabla\cdot\mathbf{F}\right)\,dV=\iint_S \left(\mathbf{F}\cdot\mathbf{n}\right) \,dS$$

with $\mathbf{n}$ being the normal vector of $\mathbf{F}$ (I'll let you see the difference with that you wrote).
In your case, having $\nabla\cdot\mathbf{F}=0$ (which means that there is no net flux), gives:
$$\iiint_V\left(\nabla\cdot\mathbf{F}\right)\,dV=\iiint_V0\cdot\,dV=0\implies\iint_S \left(\mathbf{F}\cdot\mathbf{n}\right) \,dS=0$$
Please note that:
$$\iiint_V\left(\nabla\cdot\mathbf{F}\right)\,dV\neq\left(\nabla\cdot\mathbf{F}\right)\iiint_V\,dV.$$
