Combining Steve's idea with the Beurling-Hormander uncertainty principle yields the kind of result you're looking for.

The BHUP is the following:

If $f\in L^1(\mathbb{R})$ and $ \int\int_{\mathbb{R}^2}|f(x)\hat f(y)|e^{|xy|}dxdy<\infty, $ then $f=0$.

As a corollary, we have a Hardy uncertainty principle type result:

If $\phi(x)$ and $\psi(y)$ are convex conjugates (so that $\phi(x)+\psi(y)\geq xy$), and if $\int_{-\infty}^\infty|f(x)|e^{\phi(x)}dx<\infty$ and $\int_{-\infty}^\infty |\hat f(y)|e^{\psi(y)}dy<\infty$, then $f=0$.

These theorems are stated for $\hat f(y)=\int_{-\infty}^\infty f(x)e^{-ixy}dx$. Note that $\psi(y)$ is essentially the Legendre transform of $\phi(x)$:

$\psi(y)=\sup_x xy-\phi(x)$.

Also, note that we can ignore a bounded part of the domain of $\hat f$, since $|\hat f(y)|\leq\Vert f\Vert_1$.

Now like Steve said, setting $f(x)=g(e^{-x})$, we have $\hat f(y)=Mg(iy)$. Note that $f\in L^1(\mathbb{R},dx)$ if and only if $g\in L^1(\mathbb{R}_{>0},\frac{dx}x)$. Thus the BHUP for Mellin Transforms is as follows:

BHUP for Mellin Transforms: Let $g$ be a function in $L^1(\mathbb{R}_{>0},\frac{dx}x)$.

If $ \int_0^\infty\int_{-\infty}^\infty |g(x)||Mg(iy)|e^{|\ln(x)\cdot y|}\frac{dx}{x}dy<\infty$, then $g=0$.

If $\phi(x)$ and $\psi(y)$ are functions such that $\phi(x)+\psi(y)\geq xy$, then the Hardy uncertainty principle type result translates to:

Corollary: If $\int_0^\infty |g(x)|e^{\phi(-\ln x)}\frac{dx}x<\infty$ and $\int_{-\infty}^\infty |Mg(iy)|e^{\psi(y)}dy<\infty$, then $g=0$.

Now suppose that $g$ is a non-zero function in $L^1(\mathbb{R}_{>0},\frac{dx}x)$ such that $|g(t)|=O(e^{-t^r}/t^\delta)$ for some $\delta>0$. If we choose $\phi(x)=e^{-rx}$, then

$\int_0^\infty |g(x)|e^{\phi(-\ln x)}\frac{dx}x <\infty$.

Thus $Mg(it)$ cannot decay (significantly) faster than $|t|^{-1}e^{-\psi(t)}$, where $\psi(t)$ is (something like) the Legendre transform of $e^{-rx}$. What we need is $\psi(t)\geq |xt|-e^{-rx}$ for all $t$ outside some interval. I think that taking

$\psi(t)=\frac tr -\frac tr \ln\left(\frac{-t}r\right)$ for $t<-r$

and

$\psi(t)=-\frac tr +\frac tr \ln\left(\frac{t}r\right)$ for $t>r$ works.

Thus as $t\to\infty$, $|Mg(\pm it)|$ cannot decay faster than roughly $e^{-\frac tr\ln\frac tr}$.

This isn't quite as sharp as you were hoping for, but I probably haven't done the most careful analysis.

Combining Steve's idea with the Beurling-Hormander uncertainty principle yields the kind of result you're looking for.

The BHUP is the following:

Beurling-Hormander Uncertainty Principle: If $f\in L^1(\mathbb{R})$ and $$ \int\int_{\mathbb{R}^2}|f(x)\hat f(y)|e^{|xy|}dxdy<\infty, $$ then $f=0$.

As a corollary, we have a Hardy uncertainty principle type result:

Corollary: If $\phi(x)$ and $\psi(y)$ are convex conjugates (so that $\phi(x)+\psi(y)\geq xy$), and if $$\int_{-\infty}^\infty|f(x)|e^{\phi(x)}dx<\infty\quad\mbox{and}\quad\int_{-\infty}^\infty |\hat f(y)|e^{\psi(y)}dy<\infty,$$ then $f=0$.

These theorems are stated for $\hat f(y)=\int_{-\infty}^\infty f(x)e^{-ixy}dx$. Note that $\psi(y)$ is essentially the Legendre transform of $\phi(x)$: $$\psi(y)=\sup_x xy-\phi(x).$$

Also, note that we can ignore a bounded part of the domain of $\hat f$, since $|\hat f(y)|\leq\Vert f\Vert_1$.

Now like Steve said, setting $f(x)=g(e^{-x})$, we have $\hat f(y)=Mg(iy)$. Note that $f\in L^1(\mathbb{R},dx)$ if and only if $g\in L^1(\mathbb{R}_{>0},\frac{dx}x)$. Thus the BHUP for Mellin Transforms is as follows:

BHUP for Mellin Transforms: Let $g$ be a function in $L^1(\mathbb{R}_{>0},\frac{dx}x)$.

If $$ \int_0^\infty\int_{-\infty}^\infty |g(x)||Mg(iy)|e^{|\ln(x)\cdot y|}\frac{dx}{x}dy<\infty,$$ then $g=0$.

If $\phi(x)$ and $\psi(y)$ are functions such that $\phi(x)+\psi(y)\geq xy$, then the Hardy uncertainty principle type result translates to:

Corollary: If $$\int_0^\infty |g(x)|e^{\phi(-\ln x)}\frac{dx}x<\infty\quad\mbox{and}\quad\int_{-\infty}^\infty |Mg(iy)|e^{\psi(y)}dy<\infty,$$ then $g=0$.

Now suppose that $g$ is a non-zero function in $L^1(\mathbb{R}_{>0},\frac{dx}x)$ such that $|g(t)|=O(e^{-t^r}/t^\delta)$ for some $\delta>0$. If we choose $\phi(x)=e^{-rx}$, then

$$\int_0^\infty |g(x)|e^{\phi(-\ln x)}\frac{dx}x <\infty.$$

Thus $Mg(it)$ cannot decay (significantly) faster than $|t|^{-1}e^{-\psi(t)}$, where $\psi(t)$ is (something like) the Legendre transform of $e^{-rx}$. What we need is $\psi(t)\geq |xt|-e^{-rx}$ for all $t$ outside some interval. I think that taking $$\psi(t)=\frac{|t|}r \ln\left(\frac{|t|}r\right)-\frac{|t|}r\qquad\mbox{for}\quad |t|>r$$ works.

Thus as $t\to\infty$, $|Mg(\pm it)|$ cannot decay faster than roughly $e^{-\frac tr\ln\frac tr}$.

This isn't quite as sharp as you were hoping for, but I probably haven't done the most careful analysis.