How does a tsunami affect biology and other matter in its path while in deep water
The answer to your question, based on linear theory, is no. The short answer is that the slope of these waves is very small, and the displacement of the particle paths is proportional to (ak) (times, perhaps, a depth dependent term) with a the wave amplitude and k its wavenumber. Now let's make this answer rigorous. 

Recall, for irrotational inviscid 2 dimensional water waves the governing equation is

$$\nabla^2 \phi = 0$$

where $\nabla^2 \equiv \partial_{xx}+\partial_{zz}$ and $\phi$ is the velocity potential. We require particles on the free surface to remain there, and for the pressure to be continuous across the air-water interface, which gives us (assuming the atmospheric pressure to be 0) our surface boundary conditions 

$$\left.\eta_t+\phi_x\eta_x = \phi_z\right|_{z=\eta}; \quad \quad \left. \phi_t+\frac{1}{2}(\nabla \phi)^2 +gz=0\right|_{z=\eta},$$

where $\eta(x,t)$ is the free surface displacement. Finally, we require the bottom to be impermeable, hence

$$\phi_z = 0 \quad @ z= -h$$

for $h$ the depth of the water. These are a closed set of equations. Note, the governing equation, ie Laplace's equation, is linear and hence simple to solve. However, the free surface conditions are nonlinear, and more significantly, evaluated at a moving interface. This last condition makes these equations very difficult to solve. 

Therefore, to make progress, we generally take asymptotic expansions of these equations. Here, we'll just linearize the equations about some small parameter $\epsilon = ak$ for characteristic wave amplitude $a$ and wavenumber $k$. We will also consider periodic monochromatic waves, so that one can show that the following $\eta,\phi$ satisfy the governing equations to first order in $\epsilon$: 

$$\eta = a \cos \theta; \quad \phi = a\omega \sin \theta \frac{\sinh k(z+h)}{\sinh kh},$$

where $\omega$ is the angular frequency given by the dispersion relationship

$$\omega = \sqrt{gk \tanh kh},$$
and $\theta\equiv kx-\omega t$. 

To answer your question, we would like to know the behavior of a particle with position $(\xi+x_o,\zeta+z_o)$ originally located at some position $(x_o,z_o)$.This is the Lagrangian description of the fluid. Recall, we can relate the Eulerian and Lagrangian descriptions of a fluid through the velocity field, that is 

$$\phi_x = \frac{d \xi}{d t}; \quad \phi_z = \frac{d\zeta}{dt}.$$

Substituting in the relations for $\phi$ found above, and then integrating in time, we find 

$$\xi =-a\frac{\cosh k(z_o+h)}{\sinh kh} \sin \theta_o; \quad \quad \zeta = a\frac{\sinh k(z_o+h)}{\sinh kh} \cos \theta_o,$$
where $\theta_o =kx_o - \omega t$. We note that the particle displacements from the origin scales with the wave amplitude $a$. 

Now, we can eliminate $\theta_o$ by squaring the expressions for the particle position to find 

$$\frac{\xi^2}{\left(a\frac{\cosh k(z_o+h)}{\sinh kh} \right)^2 } +\frac{\zeta^2}{\left(a\frac{\sinh k(z_o+h)}{\sinh kh} \right)^2 }= 1$$

which describes an ellipse. I've attached a figure from Kundu and Cohen which illustrates this. 

Now we can finally talk about some physics, and answer your question. For deep water waves, when $kh \ll 1$, the above relation reduces to the equation for a circle, hence the particles have a circular orbit which decays with depth (with an e-folding scale that goes like the wave number). For shallow water waves, i.e.  $kh$ small, our equation describes an ellipse with a semi major axis that is much much larger than the semi minor axis, hence there's is nearly horizontal motion that doesn't decay much with depth. 

Tsunamis have wavenumbers that are large enough so that even in the middle of the ocean they fall into the shallow water regime, meaning that the particle trajectories are approximately very flat ellipses, or horizontal lines, and this does not decay with depth. However, their deviations from their origin will be very small, as their slope $ak$ is very very small, e.g. $\mathcal{O}(1^{-5})$ for numbers I looked up online. Therefore, it's very unlikely that this will have much of an effect on a passive tracer.

Now, the story is very different as the wave approaches shore and steepens. But this is a different story for a different day. 

I hope this helps. Feel free to ask any other questions,

Nick