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Given $X$ a random variable that takes values on all of $mathbb{R}$ with associated probability density function $f$ is it true that for all $r > 0$

$$E left[ int_{X-r}^{X+r} f(x) dx right] ge E left[ int_{X-r}^{X+r} g(x) dx right]$$

for any other probability density function $g$ ?

This seems intuitively true to me and I imagine if it were to be true that it has been proven but I can't find a similar result on the standard textbooks, even a reference is welcome.

Taking the particular case of small $r$ ($r to 0$) and continuous $f$, your inequality turns equivalent to

$$ int f^2 ge int f g $$

with the restrictions $int f = int g = 1$ and $fge 0$, $gge 0$. This is clearly false. For a fixed $f$ we maximize $int f g$, not by choosing $g=f$, but by choosing $g$ concentrated around the mode (maximum) of $f$.

Incidentally, your assertion has a simple interpretation: suppose I have to guess the value of a random variable $x$ with pdf $f$, so that I win if the absolute error $e=|x- hat x|$ is less than $r$. If the inequality were true, then the conclusion would be that my best estrategy (in terms of expected win rate) is to make a random guess , by drawing my $hat x$ as an independent random variable with the same density as $x$. But this is not true, the optimal guess is to choose a deterministic value, that which maximizes the respective integral; for small $r$ this is the mode of $f$ (maximum a posteriori).

Unfortunately, your intuitive conjecture is INCORRECT.

Let $f(x)$ be the PDF of the random variable $X$ and $F(x)$ be its cumulative PDF, so that $F'(x)=f(x)$, or
$$F(x)=int_{-infty}^x f(t)dt$$
Similarly, let $g(x)$ be another PDF with cumulative PDF $G(x)$. Then the expected value of the integral
$$int_{X-r}^{X+r} g(x)dx$$
is equal to
$$int_{-infty}^infty int_{x-r}^{x+r} f(x)g(t)dtdx=int_{-infty}^infty (G(x+r)-G(x-r))f(x)dx$$
By using integration by parts, we have that
$$int_{-infty}^infty (G(x+r)-G(x-r))f(x)dx=int_{-infty}^infty (F(x+r)-F(x-r))g(x)dx$$
Consider this simple counterexample. Let $r=1$, and suppose that
$$f(x)=frac{1}{pi}frac{1}{1+x^2}$$
Then, if your conjecture is true, for no function $g$ will the integral
$$frac{1}{pi}int_{-infty}^infty (arctan(x+1)-arctan(x-1))g(x)dx$$
even surpass the value
$$frac{1}{pi^2}int_{-infty}^infty frac{arctan(x+1)-arctan(x-1)}{1+x^2}dxapprox 0.1475$$
However, suppose that we let
$$g(x)=frac{4}{pi}frac{1}{1+4x^2}$$
Then the value of our integral is equal to
$$frac{4}{pi^2}int_{-infty}^infty frac{arctan(x+1)-arctan(x-1)}{1+4x^2}dxapprox 0.3743$$
which disproves your conjecture.