In a strict use of terms, there are two different senses in which a gravitational field may be uniform, and they are not the same. (I'll describe them in a moment). But the main qualitative answer to your question is that to create the kind of field you want you would need a large mass shaped as a very wide flat plane or as a long straight cylinder (see below for the reason for two answers). The field would then cause acceleration of other objects towards that plane or cylinder. So yes, it does cause acceleration. But as usual, two objects in freefall next to one another would have small or zero acceleration relative to each other.
Now for the more technical part.
We describe gravitation by setting up a system of coordinates in some region of spacetime and then providing an equation which says how to find the spacetime interval between neighbouring events. The equation is called the line element or the metric (more strictly, the metric is the set of coefficients that appear in the equation). The line element for the uniform gravitational field has the form
$$
ds^2 = -\alpha^2(x) dt^2 + dx^2 + dy^2 + dz^2
$$
where $\alpha(x)$ is a function of $x$, and we should consider two possibilities:
$$
\alpha(x) = a x \\
\alpha(x) = c e^{k x}
$$
where $a$ and $k$ are constants and $c$ is the speed of light.
The first possibility gives
$$
ds^2 = -a^2 x^2 dt^2 + dx^2 + dy^2 + dz^2.
$$
This is called the Rindler metric. It describes an example of flat spacetime, so in one sense one might say there is no gravity here. But it describes what goes on if your system of coordinates has constant proper acceleration, like coordinates set up inside a rocket whose engine is constantly providing the same proper acceleration to the rocket. Inside such a rocket you see apples fall to the floor etc. People sometimes call this a "uniform gravitational field". But it is uniform in one sense and not in another. There are no tidal stretching or squeezing effects, which indicates a kind of uniformity, but on the other hand the acceleration due to gravity depends on $x$, so it is not uniform in that sense. For the Rindler metric one finds that the acceleration, relative to the coordinates, of an object released from rest is proportional to $1/x$. On the other hand, if you are sitting high up in the field and dangling a heavy object on a light rope, then the force you have to apply to your end of the rope does not depend on where the object is! How can these two statements not be mutually contradictory? (Exercise for the (expert) reader).
Now for the second case.
$$
ds^2 = -c^2 e^{2kx} dt^2 + dx^2 + dy^2 + dz^2.
$$
In this case, let's see what happens if we shift the origin of spatial coordinates by some constant $x_0$ and multiply the temporal coordinate by a constant factor $e^{k x_0}$.
Define
$$
X = x - x_0, \;\;\;\;\;\; T = e^{k x_0} t.
$$
One finds
$$
ds^2 = -c^2 e^{2kX} dT^2 + dX^2 + dy^2 + dz^2.
$$
In other words we have the same metric as before, just with the new coordinate labels. It follows immediately that all gravitational effects of this metric will be the same no matter where you are in spacetime. So this earns the right to be called a "uniform gravitational field". But this uniformity includes that everywhere it produces a tidal stretching/squeezing effect on any object, because the spacetime curvature is now non-zero.
But to answer your question, this second example of a "uniform gravitational field" is impossible, because no configuration of matter can give rise to it (the matter would have to have pressure but no energy density, which is not possible).
Finally, let me return to the opening paragraph. You have near Earth's surface an example of a gravitational field which is close to uniform on distance scales small compared to the radius of the Earth. That is enough to tell you that approximately uniform gravitational fields are possible, and it gives you a hint of how they may be produced.
But I get the impression from the question that what you really want to know is whether you can have a large and steady gravitational acceleration without any tidal stretching effects which might make the voyaging astronaut uncomfortable. For that you want the Rindler metric, and the configuration of matter which gives this metric to reasonable approximation (in a plane at least) is one where the gravitational acceleration falls inversely with distance. A large mass shaped as a long cylinder would do that.
But using gravity to accelerate spaceships is never the whole solution, because the spaceship will fall down towards the gravitating body. It will either crash on the body, or pass through a hole if there is one, or it will have to be steered aside. If it gets very fast then it will be difficult to steer aside, and in any case it will then slow down again as it leaves the gravitational potential well.
In view of all this, I suspect that your best bet may be to 'hitch a ride' by approaching a star which is already moving fast relative to your rocket, and then use a sling-shot effect, or just be content to orbit the star for a while if it is moving towards your destination.
