Will the sea form a concave surface? Without neglecting the fact that the earth is rotating, let us be in a ship (such that there is no relative motion between sea and ship) in a sea that surrounds the northern most or southern most point of earth's rotational axis. 
Will the ship crew see a concave surface water (using Newton's rotating bucket argument)?

(source: books24x7.com) 

(source: physicalgeography.net) 
 A: No, it won't.
In a bucket, the gravitational forces acting on each particle of the fluid are parallel (if the the bucket is small with respect to the Earth radius, which I think we can always assume).  Without rotations, the fluid will form a plane surface inside the bucket (which is neither convex nor concave).
This surface coincide with a surface of constant potential energy (given by the gravitational field of the Earth).
In the presence of a centripetal force, the surface of constant potential energy is now a concave surface, due to the presence of a potential energy term given by the centripetal field.
The gravitational forces acting on each particle of the ocean's water are not parallel (in fact the Earth is a sphere). 
The centripetal force due to Earth rotation will not make the ocean concave. The effect of centripetal forces in this case is the fact that the Earth radius is larger at the equator than at the poles. As a consequence of this, the shape of the Earth is that of an ellipsoid of revolution.
To see this more clearly, one can calculate the potential energy of a point on the surface, considering the sum of gravitational acceleration and centripetal forces.
The sea level coincide with a surface of constant potential energy.
Note that centripetal force do not only deform the shape of the waters from a perfect sphere to an ellipsoid, but they deform also the Earth crust.
A: The same effect does occur, but the magnitude is very small compared to the natural curvature of the earth.
The follow up question would be, could there be a planet on which the ocean formed a concave surface at the poles?
Intuitive answer
If the water wants the flow away from the poles to the outside so badly that it forms a concave surface it will want to flow way from the axis all that much more the further out it goes, and will thus fling the liquid off the planet (along with everything else).
Mathematical analysis
The surface of a liquid exposed to an atmosphere is at a constant pressure. Assuming the gravity due to mass of the ocean at higher altitudes than the minimum surface height is negligible, the gravity of a planet will increase pressure according to:
$$\frac {\partial P}{\partial r}=-\frac{G\,M\,\rho}{r^2}$$
Where r is the spherical coordinate corresponding to $\sqrt{r^2+z^2}$ in cylindrical coordinates. So the cylindrical pressure contributions due to gravity are:
$$\frac {\partial P}{\partial z}=-\frac{G\,M\,\rho\,z}{(r^2+z^2)^\frac32}$$
$$\frac {\partial P}{\partial r}=-\frac{G\,M\,\rho\,r}{(r^2+z^2)^\frac32}$$
And the rotation of the planet will add a partial pressure gradient:
$$\frac {\partial P}{\partial r}=\rho\,\Omega^2\,r$$
So the total pressure gradients will be:
$$\frac {\partial P}{\partial z}=-\frac{G\,M\,\rho\,z}{(r^2+z^2)^\frac32}$$
$$\frac {\partial P}{\partial r}=-\frac{G\,m\,\rho\,r}{(r^2+z^2)^\frac32}+\rho\,\Omega^2\,r$$
Since any surface would have to have a constant pressure we can state that these pressure contributions would have to cancel along the surface of the ocean.
$$\frac {\partial P}{\partial z} dz+\frac {\partial P}{\partial r} dr=0$$
$$-\frac{G\,M\,\rho\,z}{(r^2+z^2)^\frac32} dz - \frac{G\,M\,\rho\,r}{(r^2+z^2)^\frac32}dr+\rho\,\Omega^2\,r dr=0$$
$$\frac{dz}{dr}=\frac{r}{z}\left(\frac{\Omega^2}{G\,M}(r^2+z^2)^\frac32 - 1\right)$$
To examine the convexity of the surface near the poles we can look at the sign of $\frac{d^2z}{dr^2}$ for r near zero.
$$\frac{d^2z}{dr^2}=\frac{1}{z}\left(\frac{\Omega^2}{G\,M}(r^2+z^2)^\frac32 - 1\right)+\frac{3\,r^2}{z}\frac{\Omega^2}{G\,M}\sqrt{r^2+z^2}$$
$$\frac{d^2z}{dr^2}(r=0)=\frac{1}{z}\left(\frac{\Omega^2}{G\,M}z^3 - 1\right)$$
This is indicates that the pole surfaces would be concave for:
$$|z(r=0)|>\left(\frac{\Omega^2}{G\,M}\right)^\frac13$$
Unfortunately if plug this into our slope equation we'll see a problem as then $\frac{dz}{dr}$ is positive for all r which would result in an unbounded planet. While I have nothing against unbounded planets, it would invalidate the assumption that the mass from the liquid above the lowest point is negligible compared to M (the mass below the lowest point)
Surface tension?
Could surface tension hold together a droplet sized "planet" while spinning fast enough to produce a convex surface? No, even then it would be unstable: http://physics.aps.org/articles/v1/38
A: The water is not in a bucket it is free to move around. An it does move to the equator because of the earths rotation (Here also gravitational forces from the moon play a role). So no I don't think they would see a convex surface, not even at a microscopic Level. They would however see that the water surface is a tiny bit less concarve then at the equator.
A: I think (and somebody correct my math if I get this wrong), there is a tiny effect, and, no, it's not concave ever.
If the Earth was a flat rotating disc, then this would happen.  The Earth's sphere shape makes the 2 dimensional math inaccurate, but lets look at the 2-d math anyway.
Newton's bucket mathematics  
Source: https://en.wikipedia.org/wiki/Bucket_argument
Little g is gravitational acceleration (9.8 meters/second squared), and Ω is the rate of rotation, sometimes expressed as ω.   1 rotation in 24 hours is 1/(24*60*60) rotations per second, or 1/86400 and r is . . . well, that's where the equation breaks down, cause this equation is for a flat spinning container of water, not a globe, but we can use 2 definitions of r, radius is one and distance, pole to equator, or 1/4 circumference is the other.
radius of the earth = 6,378,100 meters, so we have 1/19.6 * (6,378,100/86400)^2 = (73.8)^2/19.6 = 278 meters, which is how much higher the edges of the ocean would be if the earth was a flat spinning disk of water, at current rotational speed, at 1 earth radius.
Now if we use 1/4 circumference, a rather neat 10 million as an approximation, you work out the math and get 683 meters.    Needless to say, neither is a very good approximation of the earth's rotational force because the earth's bulge at the equator is some 42,700 meters greater than at the poles, but then, we're using 2 dimensional math for a 3 dimensional approximation so we shouldn't be shocked by missing rather badly.
If we, just for fun, calculate this for 1 second or arc-second (angle, not time), that's 1/3,600 of 1 degree, or about 30.8 meters.   Using the bucket equation, (30.8/86400)^2 / 19.6 and you get 1/15.4 million meters, give or take, rise due to the spinning of the earth - virtually nothing, but you have to remember, 1 rotation every 24 hours is very slow, so this small a number shouldn't be surprising.
Now if we calculate the curvature of the earth for 1 second of arc, and the expected drop off, that's simple enough to approximate using the Pythagorean theorem, so,
Radius = 6,378,100 meters
1 arc second in distance, about 30.8 meters
Apply A^2 + B^2 = C^2 or in this case A = (C^2 - B^2) ^ 1/2 and you get A = (40680159610000 - 949) ^ 1/2 = 6378099.999925 meters (or roughly 1/13,000 of a meter in drop off due to curvature) at 1 arc second.
So, you get about 1,000 times more effect from curvature as you get from the earth's rotation, even on a relatively small scale.   Now, if the earth rotated faster, like 35 times faster or, something, then these 2 forces would balance each other out and you'd get flat water on the poles.    That kind of rotation speed, would of-course, be impossible cause at that speed the equatorial velocity would be greater than the escape velocity and the Earth would fly apart.   
I think this is a fun mathematical exercise and I'll try to look more closely later and see if there is a more precise relation, but I'm pretty sure it's impossible to have any concavity, even for a very fast spinning object.
Maybe a Kerr black hole, but . . . don't take my word for it.   I don't understand those very well.
This was a curious mathematical exercise, and if my calculations need fixing, I invite correction. 
