Nitsche worked out.
$$ \langle\mathbf{u}_t \cdot \mathbf{v} \rangle + \langle\nabla \eta_0 \cdot \mathbf{v} \rangle + \frac{\epsilon}{3}h^2(\langle \nabla\cdot \mathbf{u}_t \nabla\cdot \mathbf{v}\rangle - \int_{\partial \Omega} \mathbf{v} \cdot \mathbf{n}\nabla \cdot \mathbf{u}_t ~\mathrm{ds})= 0$$
with Dirichlet BC expressed as $\nabla \cdot \mathbf{u}_t +\kappa \mathbf{u}_t \cdot \mathbf{n} = 0$ and substitute into the expression above, so that the last term looks like
$$\int_{\partial \Omega} \mathbf{v} \cdot \mathbf{n} \kappa \mathbf{u}_t \cdot \mathbf{n}~\mathrm{ds}$$
where $\kappa$ is a large enough constant.
F1 = Constant(1/dt)*dot((u-u0),v)*dx \
+ eps/(Constant(3)*dt)*h*h*inner(div(u-u0),div(v))*dx \
+ dot(grad(eta0),v)*dx
F1 += -Constant(kappa)*eps/Constant(3*dt)*dot(v,n)*(dot(u0,n)-dot(u,n))*ds
F2 = Constant(1/dt)*(eta-eta0)*q*dx - h*dot(u0,grad(q))*dx