Relation between strain and velocity The strain tensor writes $\epsilon_{ij}=\frac{1}{2}\Big(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_i}{\partial x_j}\Big)$ with $u_i$ the displacement in the $i$ direction. 
Then $\frac{\partial \epsilon_{ij}}{\partial t}=\frac{1}{2}\Big(\frac{\partial }{\partial x_j}\frac{ \partial u_i}{\partial t}+\frac{\partial }{\partial x_i}\frac{ \partial u_j}{\partial t}\Big)$. 
Is it right to identify $\frac{ \partial u_i}{\partial t}$ with the velocity $v_{i}$ of the particles of the material, thus yielding the tensorial strain-velocity relation : $\frac{\partial \mathbf{\epsilon}}{\partial t}=\frac{1}{2}(\nabla  \mathbf{v}+(\nabla  \mathbf{v})^T)$ ? 
 A: Actually your equation is pretty close.  The difference in velocity between two neighboring material points in a deforming fluid is given by $$\Delta \mathbf{v}=(\nabla \mathbf{v})^T\centerdot d\mathbf{s}$$ where $d\mathbf{s}$ is the differential position vector between the two material points and $(\nabla \mathbf{v})^T$ is the transpose of the velocity gradient tensor.  The rate of strain tensor (which factors out the effect of rotation of the fluid parcels) is given by:
$$\mathbf{E}=\frac{[(\nabla \mathbf{v})+(\nabla \mathbf{v})^T]}{2}$$such that $$\frac{D(ds)^2}{Dt}=d\mathbf{s}\centerdot \mathbf{E} \centerdot d\mathbf{s}$$where D/Dt is the material derivative.
A: your question: "Is it right to identify $\frac{\partial u_i}{\partial t}$ with the velocity vi of the particles of the material,.."
The correct formula implies material derivatives (as explained in the former answer) so the description must be in Lagrangian or material coordinates so that
$$\frac{\partial}{\partial t}\equiv\frac{\partial\vec{u}}{\partial t} + \vec{v}\cdot\nabla\vec{v}$$
