No, such functions do not exist. Let $u=\log|f|$. Your condition implies that $u(x)\leq O(|x|^\rho)$,
where $0\leq \rho<1$, so the Poisson integral of $u^+$ is convergent,
and one can obtain the formula
$$u(z)=\frac{y}{\pi}\int_{-\infty}^\infty u(t)\frac{dt}{(t-x)^2+y^2}\quad\quad +ky+\log|B(z
)|,$$
where $z=x+iy$, and $B$ is a Blaschke product for the upper half-plane.
See, for example, theorems 7 and 5 in Levin, Distribution of
zeros of entire functions, Chap V, sect 3. (Such functions are called of class $A$ in Levin's books,
and of Cartwright's class in other books. It is a rather deep result of Cartwright that
convergence of the Poisson integral of $u^+$ implies in this case the convergence of
the Poisson integral of $u$).
Now the condition that $f$ is of zero type implies that $k=0$.
(See Thm 6 in the same book, same chapter).
Then $u$ is estimated from above by the Poisson integral since $\log|B|\leq 0$.
Inserting to the Poisoon integral $|x|^\rho$ instead of $u(x)$, we
obtain an upper estimate for $u$. This estimate involves an integral which can be
computed, and computation shows that $u$
must be of order $\rho<1$.
So it cannot be of order $1$, minimal type.
On your other question (explicit examples of functions of order $1$ and type zero), of course
the simplest answer is an infinite product, take zeros at the points $\pm [r\log r]$, for example,
where $[.]$ is the integer part. If you don't like products then you can construct examples
with explicit power series. Take the coefficients $a_n=n^{-n}(\log n)^{-n}$.
Hope this is explicit enough.
EDIT. Here is another way to construct a function of minimal type order $1$.
Functions $f$ of at most minimal type order 1 have the following description
$$f(z)=\frac{1}{2\pi i}\; {\mathrm{res}}_{\zeta=0} F(1/\zeta)e^{z\zeta},$$
where $F$ is an arbitrary entire function with the property $F(0)=0$. This is called Leau's theorem
(19 century).
Now if you want $f$ to be exactly of order $1$ (not "at most"), take $F$ of infinite order.
For example $F(z)=\exp(\exp(z))$. So you have an explicit formula involving an integral.
