Let's assume that the cow is spherical .. sorry, assume that the walker is a box with height $h$, width $w$ and length $l$. The box needs to cover a horizontal distance $D$ while being hit by a minimal amount of water.
Further assume that the air is uniformly filled with $\rho$ kilograms of water per cubic meter, moving straight downwards at a uniform speed $v_r$ (since we imagine that the raindrops have reached terminal velocity).
If the box is walking at speed $v_h$, how much water hits him? In each infinitesimal moment $\Delta t$, the water in the air falls a vertical distance of $\Delta t v_r$, and $\Delta t v_r w l\rho$ amount of water hits the top of the box. The total amount of water to hit the top during the trip is $\frac{D}{v_h} v_r w l \rho$.
Similarly, during each infinitesimal moment $\Delta t$ the front of the box pushes into a volume $\Delta t v_h h w$ of rain-filled air. The rain is moving downwards, but hits the box nevertheless, so the total amount of water to hit the front is $\frac{D}{v_h} v_h h w \rho = D h w \rho$ -- in other words this amount is independent of $v_h$. The only influence $v_h$ has is that the larger $v_h$ is, the less water hits the top of the box. So under the above simplifying assumptions one should indeed attempt to maximize speed.
Things get rather more interesting if the rain is falling at an angle -- then the amount of rail to hit the front/back of the box does depend on $v_h$ and matters are not so simple anymore. If I remember correctly, then if there is a tailwind, there's an optimal walking/running speed above which the amount of water to hit the front increases with speed faster than the amount of water to hit the top decreases. (But I don't have time to derive that right now).
More specifically, if there's so much tailwind that, standing still, you would receive more water on your back than on your top, then you should run exactly at the wind speed in order to minimize the amount of water hitting you.