Let $\mu$ be some ergodic measure of our compact Riemannian manifold $M$, which is preserved by $f\in Diff^{1+\beta}(M)$. Is it possible that all the Lyapunov exponents of $\mu$ will be positive? Intuitively this seems wrong, but I couldn't find any general proof without assuming that $h_\mu(f)>0$ (which I don't want to restrict myself to doing).

As Will shows, the case in which $\mu$ is absolutely continuous with respect to Lebesgue measure and has density bounded away from zero and infinity is constrained in that the Lyapunov exponents of $\mu$ must sum to zero. If $\mu$ is an arbitrary ergodic measure then the Lyapunov exponents can all be positive, for example if $\mu$ is the Dirac measure on a repelling fixed point. 


Yes, because the system is conservative, the sum of the Lyapunov exponents is $0$, so they cannot all be positive. Observe that the sum of the Lyapunov exponents is $\lim_{n \to \infty} \frac{1}{n} \log \left\det \frac{ df^n}{dx}(x)\right$. But as $\mu$ is invariant under $f^n$, $\det \frac{ df^n}{dx}$ is the ratio of the density of $\mu$ at $x$ to the density of $\mu$ at $f^n(x)$. Because the density is continuous function on a compact manifold, it is bounded, so the limit is zero. 

