Short answer: Yes, the set of roots can change. $f(x)=x^2+1$ has no roots in $\mathbb{Z}$, but (the image of $f(x)$ under reduction modulo $p$) has roots in $\mathbb{F}_p$ if $p=2$ or $p\equiv 1 \pmod{4}$.
Longer answer: First, you have to be careful. If $f(x)$ is a polynomial with integer coefficients, then it is not literally a polynomial with coefficients in $\mathbb{F}_p$. Rather, the natural map "reduction modulo $p$", which is a ring map from $\mathbb{Z}$ to $\mathbb{F}_p$, induces a map from $\mathbb{Z}[x]$ to $\mathbb{F}_p[x]$, and you can consider the image of $f(x)$ in $\mathbb{F}_p[x]$.
Under this map, if $a\in\mathbb{Z}$ is a root of $f(x)$, then $a\bmod p$ is a root of $f(x)\bmod p$ in $\mathbb{F}_p[x]$. But the converse certainly does not hold; for one thing, there are infinitely many integers that are preimages of $a\bmod p$. And for another, no: you can have $f(x)\bmod p$ have roots in $\mathbb{F}_p[x]$, but no roots in $\mathbb{Z}[x]$, or only some roots. For example, for odd primes $p$, $f(x) = x^{p}-x$ has only two roots in $\mathbb{Z}$ ($x=0$ and $x=1$), but (its image under reduction modulo $p$) has $p$ roots in $\mathbb{F}_p$. So roots in $\mathbb{Z}$ yield roots in $\mathbb{F}_p$ (via reduction modulo $p$), but not conversely. Another example: $x^2+1$ has no roots in $\mathbb{Z}$, but (its reduction modulo $p$) has roots in $\mathbb{F}_p$ for every $p$ that is not congruent to $3$ modulo $4$.
More generally, the universal property of the polynomial ring gives easily:
Theorem. Let $R$ and $S$ be commutative rings. If $h\colon R\to S$ is a ring homomorphism, then $h$ induces a ring homomorphism $\overline{h}\colon R[x]\to S[x]$ of polynomial rings by $\overline{h}(r_0+r_1x + \cdots + r_nx^n) = h(r_0) + h(r_1)x + \cdots + h(r_n)x^n.$ Moreover, the map $\overline{h}$ commutes with evaluation maps in the following sense: if $\mathrm{eval}_a\colon R[x]\to R$ is the map $\mathrm{eval}_a(f(x)) = f(a)$, and $\mathrm{eval}_{h(a)}\colon S[x]\to S$ is the evaluation map at $h(a)$, then we have a commutative diagram $\begin{array}{rcl} R[x] & \stackrel{\overline{h}}{\longmapsto} & S[x]\\ &&\\ \mathrm{eval}_a\downarrow & & \downarrow \mathrm{eval}_{h(a)}\\ R &\stackrel{h}{\longmapsto} & S\end{array}$ for all $a\in R$.
In particular, if given $f\in R[x]$ and $g\in S[x]$ we let $\mathrm{Roots}_R(f) = \{r\in R\mid f(r)=0\}$ and $\mathrm{Roots}_S(g) = \{s \in S\mid g(s)=0\},$ then $h\left(\mathrm{Roots}_R(f)\right) \subseteq \mathrm{Roots}_S(\overline{h}(f)),$ but the inclusion may be proper.
For examples with proper inclusion in the setting of reduction modulo $p$, consider $f(x)=x^2+1\in\mathbb{Z}[x]$ which has not roots, but whose reduction modulo $p$, with $p=2$ or $p\equiv 1\pmod{4}$ has roots. Or $f(x)=x^p-x$, $p$ an odd prime, which has two roots in $\mathbb{Z}$ but its reduction modulo $p$ has $p$ roots in $\mathbb{F}_p$ (by Fermat's Little Theorem). For examples of proper inclusion with fields and $h$ an inclusion map, take $x^2+1\in\mathbb{R}[x]$, and consider its image in $\mathbb{C}[x]$; $x^2+1$ has no real roots, but its image in $\mathbb{C}[x]$ has roots.
Added. When dealing with inclusion, it usually makes more sense to consider the set of roots in the larger field, and just ask which ones lie in the smaller field. So we think of $x^2+1$ as a polynomial with complex coefficients, and then ask what $\mathrm{Roots}_{\mathbb{C}}(x^2+1)\cap\mathbb{R}$ is (this will equal $\mathrm{Roots}_{\mathbb{R}}(x^2+1)$).
Also, if $h(a)$, the image of $a$ is a root of $\overline{h}(f)$, this does not imply, in and of itself, that $a$ is a root of $f(x)$ (though it does if $h$ is one-to-one).
To summarize the longer answer: if $h\colon R\to S$ is a ring homomorphism, then the image of any root of $f(x)$ is a root of the image of $f(x)$, but there may be roots of the image that are not images of roots; and not everything in $R$ whose image is a root of the image of $f(x)$ is a root of $f(x)$ in general. But if $h$ is one-to-one, then $h(a)$ is a root of $\overline{h}(f(x))$ if and only if $a$ is a root of $f(x)$.