Do balls of finite radius have finite volume?

Consider the open unit ball in $\mathbb{R}^2$ with polar coordinates $(r,\theta)$ and a metric of the form $g=dr^2+f^2(r)d\theta^2$, where $f$ is a smooth positive function chosen so that $g$ is well behaved at the origin. The set $B_r(0)$ is the same as in the Euclidean metric, since radial geodesics are unchanged, but the volume differs; a simple computations shows $$ \operatorname{vol}(B_r(0))=2\pi\int_0^{\min(1,r)}f(\rho)d\rho $$ Thus $\operatorname{vol}(B_1(0))$ can be made to diverge by choosing a suitable $f$ which diverges as $r\to 1^-$.

I haven't computed it, but I suspect the curvature of such a metric also diverges as $r\to 1^-$, and that this specific behavior can be controlled by a lower bound on the curvature.

Edit:

Without any kind of completeness assumptions, one can construct even more pathological counterexamples. Here's one which is flat and simply connected.

Let $S$ be the universal cover of the punctured Euclidean plane the with projection $\pi:S\to\mathbb{R}^2\setminus\{0\}$ (equipped with the pullback metric), and let $\rho:S\to\mathbb{R}$ be the "radial" function defined by $\rho(p)=\|\pi(p)\|$. The set $\{p\in S:\rho(p)<r\}$ is an open submanifold of infinite volume, since it is an infinite-sheeted cover of a punctured open ball. Additionally, one can show by constructing paths which loop arbitrarily tightly around the origin that $d(p,q)\le\rho(p)+\rho(q)$. Thus, any ball $B_r(p)$ in $S$ with $r>\rho(p)$ has infinite volume.

With this kind of local pathological behavior in mind, I don't think the size of geodesic balls can be controlled by bounding the curvature alone.