$\newcommand\si\sigma\newcommand\om\omega\newcommand\z{\mathbf z}\newcommand\R{\mathbb R}$Let us assume that your $d\Omega$ means $d\z$, so that $$I_\si(s):=\int_{\R^k} p_{Z}(z-s) f(z)\, dz,$$ and we shall assume that this definition holds for all $s\in\R^k$.

So, $$I_\si(s)=Ef(s+\si Z),$$ where $Z$ is a standard normal random vector in $\R^k$.

Let $\si\downarrow0$. Then for each $s\in\R^k$ we have $s+\si Z\to s$ in distribution, and hence $I_\si(s)=Ef(s+\si Z)\to f(s)$ for any bounded continuous function $f$.

Suppose now that for some bounded function $\om\colon[0,\infty)\to[0,\infty)$ we have $\om(0+)=0$ and $|f(x)-f(y)|\le\om(h)$ for any $x$ and $y$ in $\R^k$ such that $|x-y|\le h$; so, $\om$ is a modulus of continuity of $f$; here $|\cdot|$ denotes the Euclidean norm on $\R^k$ and on $\R$. (In particular, $f$ has such a modulus of continuity if $f$ is a continuous function with a compact support, or if $f$ is bounded and Hölder continuous.) Then for all $s\in\R^k$ $$|I_\si(s)-f(s)|=|Ef(s+\si Z)-f(s)|\le E\om(\si Z)\to\om(0+)=0,$$ by dominated convergence, since $\si Z\to0$ in probability. It follows that, whenever a bounded modulus of continuity exists, $I_\si(s)\to f(s)$ uniformly in all $s\in\R^k$.

Suppose now that $f$ is twice differentiable, with the second derivative bounded from above in the sense that for some real $b$ and all $s$ and $z$ in $\R^k$ we have $f''(s)(z,z)\le b|z|^2$. Then $$f(s+\si Z)-f(s)\le\si f'(s)(Z)+ b\si^2|Z|^2.$$ Taking now the expectations, we get $$I_\si(s)-f(s)=Ef(s+\si Z)-f(s)\le bn\si^2.$$ Similarly, if $f$ is twice differentiable, with the second derivative bounded from below in the sense that for some real $a$ and all $s$ and $z$ in $\R^k$ we have $f''(s)(z,z)\ge a|z|^2$, then $$I_\si(s)-f(s)\ge an\si^2.$$

$\newcommand\si\sigma\newcommand\om\omega\newcommand\z{\mathbf z}\newcommand\R{\mathbb R}$Let us assume that your $d\Omega$ means $d\z$, so that $$I_\si(s):=\int_{\R^k} p_{Z}(z-s) f(z)\, dz,$$ and we shall assume that this definition holds for all $s\in\R^k$.

So, $$I_\si(s)=Ef(s+\si Z),$$ where $Z$ is a standard normal random vector in $\R^k$.

Let $\si\downarrow0$. Then for each $s\in\R^k$ we have $s+\si Z\to s$ in distribution, and hence $I_\si(s)=Ef(s+\si Z)\to f(s)$ for any bounded continuous function $f$.

Suppose now that for some bounded function $\om\colon[0,\infty)\to[0,\infty)$ we have $\om(0+)=0$ and $|f(x)-f(y)|\le\om(h)$ for any $x$ and $y$ in $\R^k$ such that $|x-y|\le h$; so, $\om$ is a modulus of continuity of $f$; here $|\cdot|$ denotes the Euclidean norm on $\R^k$ and on $\R$. (In particular, $f$ has such a modulus of continuity if $f$ is a continuous function with a compact support, or if $f$ is bounded and Hölder continuous.) Then for all $s\in\R^k$ $$|I_\si(s)-f(s)|=|Ef(s+\si Z)-f(s)|\le E\om(\si Z)\to\om(0+)=0,$$ by dominated convergence, since $\si Z\to0$ in probability. It follows that, whenever a bounded modulus of continuity exists, $I_\si(s)\to f(s)$ uniformly in all $s\in\R^k$.

Suppose now that $f$ is twice differentiable, with the second derivative bounded from above in the sense that for some real $b$ and all $s$ and $z$ in $\R^k$ we have $f''(s)(z,z)\le b|z|^2$. Then $$f(s+\si Z)-f(s)\le\si f'(s)(Z)+ b\si^2|Z|^2/2.$$ Taking now the expectations, we get $$I_\si(s)-f(s)=Ef(s+\si Z)-f(s)\le bk\si^2/2.$$ Similarly, if $f$ is twice differentiable, with the second derivative bounded from below in the sense that for some real $a$ and all $s$ and $z$ in $\R^k$ we have $f''(s)(z,z)\ge a|z|^2$, then $$I_\si(s)-f(s)\ge ak\si^2/2.$$