If you assume that $M$ is oriented, then up to a multiple $I_1$ and $I_2$ are the usual integral. In this case $\partial M$ is oriented, and since this is a manifold without boundary, the integral induces a linear isomorphism $\Omega^{m-1}(\partial M)/d(\Omega^{m-2}(\partial M))\to\mathbb R$. By definition, you must have $I_2(d\beta)=0$ for any $\beta$, so up to a multiple, $I_2$ is just the usual integral. By Stokes, $I_1$ has to coincide with the same multiple of the integral on exact $m$-form. For a manifold with boundary $H^m(M)=0$, so any $m$-form is exact.

If you assume that $M$ is oriented, then up to a multiple $I_1$ and $I_2$ are the usual integral. In this case $\partial M$ is oriented, and since this is a manifold without boundary, the integral induces a linear isomorphism $\Omega^{m-1}(\partial M)/d(\Omega^{m-2}(\partial M))\to\mathbb R$. Now for any $\beta\in\Omega^{m-2}(\partial M)$ there exists an extension $\tilde\beta\in\Omega^{m-2}(M)$ and by naturality of $d$, we get $d\tilde\beta|_{\partial M}=d\beta$. Since $0=d(d\tilde\beta)$ the defining equation tells you that $0=I_1(d(d\tilde\beta))=I_2(d\beta)$. Thus you see that $I_2$ factorizes to the quotient $\Omega^{m-1}(\partial M)/d(\Omega^{m-2}(\partial M)$ and hence that is a number $a\in\mathbb R$ such that $I_2(\alpha)=a\int_{\partial M}\alpha$ for all $\alpha\in \Omega^{m-1}(\partial M)$.

Now consider $\alpha\in\Omega^{m-1}(M)$ and form $I_1(d\alpha)-a\int_M(d\alpha)$. By your defining equation the first term gives $I_2(\alpha)=a\int_{\partial M}\alpha$, so by Stokes $I_1-a\int_M$ vanishes on any exact form. But on a manifold with boundary any top degree form is exact, so $I_1=a\int_M$.