I think that the usual proof goes through in this case, although, obviously, you don't get a diffeomorphism (i.e., $C^\infty$ invertible map) identifying the two volume forms, just a just a $C^{1+\alpha}$ map with a $C^{1+\alpha}$ inverse.

Look at the steps: First, you need to find an $(n-1)$-form $\phi$ such that $\omega_2-\omega_1 = \mathrm{d}\phi$, and you should make sure that it is at least $C^{1+\alpha}$. To do this, you note that $\omega_2-\omega_1$ is zero in deRham cohomology (this is the 'equal mass' hypothesis; of course, you need to assume that $M$ is connected for this to work, but that's part of the hypothesis anyway), and then use, say, a Green's operator (which, at least, doesn't decrease regularity) for some smooth metric to write $$ \omega_2-\omega_1 = \mathrm{d}\left(G(\omega_2{-}\omega_1)\right), $$ then take $\phi = G(\omega_2{-}\omega_1)$. Second, on $M\times [0,1]$ (with $t$ as the coordinate on the second factor, consider the $n$-form (which is $C^{1+\alpha}$) $$ \omega = (1{-}t)\,\omega_1 + t\,\omega_2 + \mathrm{d}t\wedge\phi. $$ This form satisfies $d\omega = 0$ by construction, and it is never vanishing since $\omega_1$ and $\omega_2$ determine the same orientation of $M$. Third, there is a unique vector field $X$ on $M\times[0,1]$ that satisfies $$ \iota_X\left(\mathrm{d}t\wedge\omega\right) = \omega, $$ where $\iota_X$ means interior product with $X$. This vector field satisfies $\mathrm{d}t(X) \equiv 1$, so we can look at the time $1$ flow of this vector field, which carries $M\times\{0\}$ to $M\times\{1\}$. Fourth, since $\omega$ is closed and since $\iota_X(\omega) = 0$, it follows from Cartan's formula that the Lie derivative of $\omega$ with respect to $X$ is zero, i.e., that the flow of $X$ preserves $\omega$.

But now, the time $1$ flow of $X$ (which is a $C^{1+\alpha}$ vector field) is then a $C^{1+\alpha}$ map (with $C^{1+\alpha}$ inverse) from $M$ to $M$ that pulls back $\omega_2$ to $\omega_1$. This is because $\omega$ pulls back to $M\times\{0\}$ to be $\omega_1$ and it pulls back to $M\times\{1\}$ to be $\omega_2$.

I think that the usual proof goes through in this case, although, obviously, you don't get a diffeomorphism (i.e., $C^\infty$ invertible map) identifying the two volume forms, just a $C^{1+\alpha}$ map with a $C^{1+\alpha}$ inverse.

Look at the steps: First, you need to find an $(n-1)$-form $\phi$ such that $\omega_2-\omega_1 = \mathrm{d}\phi$, and you should make sure that it is at least $C^{1+\alpha}$. To do this, you note that $\omega_2-\omega_1$ is zero in deRham cohomology (this is the 'equal mass' hypothesis; of course, you need to assume that $M$ is connected for this to work, but that's part of the hypothesis anyway), and then use, say, a Green's operator (which, at least, doesn't decrease regularity) for some smooth metric to write $$ \omega_2-\omega_1 = \mathrm{d}\left(G(\omega_2{-}\omega_1)\right), $$ then take $\phi = G(\omega_2{-}\omega_1)$. Second, on $M\times [0,1]$ (with $t$ as the coordinate on the second factor), consider the $n$-form (which is $C^{1+\alpha}$) $$ \omega = (1{-}t)\,\omega_1 + t\,\omega_2 + \mathrm{d}t\wedge\phi. $$ This form satisfies $\mathrm{d}\omega = 0$ by construction, and it is never vanishing since $\omega_1$ and $\omega_2$ determine the same orientation of $M$. Third, there is a unique vector field $X$ on $M\times[0,1]$ that satisfies $$ \iota_X\left(\mathrm{d}t\wedge\omega\right) = \omega, $$ where $\iota_X$ means interior product with $X$. This vector field satisfies $\mathrm{d}t(X) \equiv 1$, so we can look at the time $1$ flow of this vector field, which carries $M\times\{0\}$ to $M\times\{1\}$. Fourth, since $\omega$ is closed and since $\iota_X(\omega) = 0$, it follows from Cartan's formula that the Lie derivative of $\omega$ with respect to $X$ is zero, i.e., that the flow of $X$ preserves $\omega$.

But now, the time $1$ flow of $X$ (which is a $C^{1+\alpha}$ vector field) is then a $C^{1+\alpha}$ map (with $C^{1+\alpha}$ inverse) from $M$ to $M$ that pulls back $\omega_2$ to $\omega_1$. This is because $\omega$ pulls back to $M\times\{0\}$ to be $\omega_1$ and it pulls back to $M\times\{1\}$ to be $\omega_2$.