Particles/solutes are always diffusing in all directions. The question to ask is, "are they diffusing fast enough to overcome being swept along by the flow?"
Short Answer: No. Don't try it, though.
Long Answer:
One governing equation that can be used to answer this question is the following:
$$ \overline v!\left(\vec r\right)\cdot \overline\nabla C_i!\left(\vec r\right) = D_i\nabla^2C_i!\left(\vec r\right)\tag{1} $$
This is the (simplifed) steady-state mass transport balance for species $i$ at a given point $\vec r$ in space. The left-hand side represents transport due to convection; the right-hand side encapsulates transport due to diffusion. $\overline v!\left(\vec r\right)$ is the fluid flow velocity, $C_i!\left(\vec r\right)$ is the concentration, and $D_i$ is the diffusivity. For concision, the explicit dependence on $\vec r$ will be omitted below.
Given that we're most interested in the transport in the direction of the flow of the tea, if we approximate the stream as a cylinder of uniformly flowing liquid with velocity $v_z$ and assume that concentration variations perpendicular to the flow are unimportant, Eq. $\left(1\right)$ becomes:
$$ v_z{dC_i\over dz} = D_i{d^2C_i\over dz^2} \tag{2} $$
if we choose the $z$-axis as lying along the stream of tea.
The most useful way to use Eq. $\left(2\right)$ is to non-dimensionalize it. That is, to re-write it so that all of the variables and coefficients have no units. For this problem, $z$ and $C_i$ are the variables to be non-dimensionalized, whereas $v_z$ and $D_i$ are parameters to be grouped into non-dimensional groups. 707
So, let's define:
$$ \eta = {z\over L}; \Theta = {C_i\over C_i^*} \tag{3} $$
$L$ is some length scale representative of the system of interest. Here, the best choice is probably the height you're holding the cup over the surface of the toxic waste. $C_i^$ is a representative concentration of a chemical species of interest, such as the nastiest player in that concentrated toxic waste. To substitute these definitions into Eq. $\left(2\right)$, I'll rewrite them as $z=\eta L$ and $C_i = \Theta C_i^$:
$$ {v_zC_i^\over L}{d\Theta\over d\eta} = {D_iC_i^\over L^2}{d^2\Theta\over d\eta^2} \tag{4} $$
The terms with derivatives in Eq. $\left(4\right)$ are now dimensionless. To make the coefficients non-dimensionless, divide both sides by $D_iC_i^*/L^2$:
$$ {v_zL\over D_i}{d\Theta\over d\eta} = {d^2\Theta\over d\eta^2} \tag{5} $$
The new grouping of parameters in Eq. $\left(5\right)$ has a special name: the Péclet number, abbreviated $\mathrm{Pe}$.
$$ \mathrm{Pe}{d\Theta\over d\eta} = {d^2\Theta\over d\eta^2} \tag{6} $$
The magnitude of $\mathrm{Pe}$ indicates the relative importance of convection relative to diffusion in a problem like this one. If the Péclet number is large $\left(\mathrm{Pe}\gg 1\right)$, it means that convection dominates the mass transfer behavior; if it's small $\left(\mathrm{Pe}\ll 1\right)$, then diffusion is the key phenomenon. If it's in the vicinity of unity $\left(\mathrm{Pe} \sim 1\right.$, or approximately $\left.0.3\leq \mathrm{Pe}\leq 3\right)$, then both diffusion and convection must be considered together.
So, can pouring tea into toxic waste poison your tea? 707