Suppose $E(X)=E(Y)$ and $\operatorname{Var}(X)>\operatorname{Var}(Y)$, do we have $E(f(X))>E(f(Y))$ for convex functions $f$?

No, variance is not strong enough to ensure such a property. Think of random variables $X$ and $Y$ with

$$ P(X=-1) = P(X=1) = 1/2 $$

and

$$ P(Y = -10)=P(Y = 10)= 1/1000, \quad P(Y = 0)= 1 - 2/1000.$$

You can check that the variance of $X$ is larger than that of $Y$, but $E[X^{10}] < E[Y^{10}]$ for instance.

Taking $f$ as the identiy function (which is convex and increasing) would yield $E(X)>E(Y)$, so there's a very fundamental contradiction in this statement.