The correct formula is, for $\chi$ primitive of conductor $q$,
$$ \sum_{n\in\mathbb{Z}}\chi(n)f\left(\frac{nx}{\sqrt{q}}\right) =
\frac{A}{x}\sum_{n\in\mathbb{Z}}\bar\chi(n)\hat f\left(\frac{n/x}{\sqrt{q}}\right). $$
Here $A:=\sqrt{q}/\tau(\bar\chi)$ is the so-called root number, it is of modulus $1$.
This formula is essentially equivalent to the functional equation of $L(s,\chi)$, but let me provide a direct proof. We start from the well-known formula
$$ \chi(n) = \frac{1}{\tau(\bar\chi)}\sum_{m=1}^q\bar\chi(m)e\left(\frac{mn}{q}\right), $$
where $e(x)$ abbreviates $e^{2\pi i x}$. See
Davenport: Multiplicative number theory, Chapter IX, equation (6).
Then we can rewrite the left hand side in the formula as
$$ \sum_{n\in\mathbb{Z}}\chi(n)f\left(\frac{nx}{\sqrt{q}}\right) =
\frac{1}{\tau(\bar\chi)}\sum_{m=1}^q\bar\chi(m) \sum_{n\in\mathbb{Z}}
e\left(\frac{mn}{q}\right) f\left(\frac{nx}{\sqrt{q}}\right).$$
Applying the Poission summation formula for the inner sum on the right hand side,
$$ \sum_{n\in\mathbb{Z}}\chi(n)f\left(\frac{nx}{\sqrt{q}}\right) =
\frac{1}{\tau(\bar\chi)}\sum_{m=1}^q\bar\chi(m)
\sum_{k\in\mathbb{Z}}\int_{-\infty}^\infty e\left(\frac{mt}{q}+kt\right) f\left(\frac{tx}{\sqrt{q}}\right)\,dt. $$
Denoting $n:=m+qk$ on the right hand side, we obtain
$$ \sum_{n\in\mathbb{Z}}\chi(n)f\left(\frac{nx}{\sqrt{q}}\right) =
\frac{1}{\tau(\bar\chi)}\sum_{n\in\mathbb{Z}}\bar\chi(n)
\int_{-\infty}^\infty e\left(\frac{nt}{q}\right) f\left(\frac{tx}{\sqrt{q}}\right)\,dt. $$
By a change of variable,
$$ \sum_{n\in\mathbb{Z}}\chi(n)f\left(\frac{nx}{\sqrt{q}}\right) =
\frac{A}{x}\sum_{n\in\mathbb{Z}}\bar\chi(n)
\int_{-\infty}^\infty e\left(\frac{nt/x}{\sqrt{q}}\right) f(t)\,dt, $$
where $A$ is as above. This is the stated (corrected) formula.
Remark of 10/14/2013. The formula as proved above holds for integrable continuous functions satisfying $|f(t)|+|\hat f(t)|\ll (1+|t|)^{-1-\delta}$ for some $\delta>0$. I had an Addendum of 10/12/2013 here with a seemingly more relaxed condition for $q>1$, but this condition turned out to be equivalent to the original one by basic facts on the Fourier transform.
Switching from comment to answer because the comment thread is getting too long.
Let $\# \mathbb{F}=q, t=q^{-s}$ and consider $L(t,\chi)$. Then (by taking the logarithmic derivative of the Euler product)
$$L(t,\chi) = \exp (\sum_{n=1}^{\infty} S_n t^n/n )$$
where $S_n = \sum_{\deg P | n} \chi(P)\deg P$
and $P$ runs through irreducible polynomials of $\mathbb{F}[T]$.
Then for any integer $d>0$,
$$\prod_{\zeta^d =1} L(\zeta t,\chi) = \exp (\sum_{n=1}^{\infty} S_{dn} t^{dn}/n )$$
Now, if $Q$ is an irreducible polynomial of degree $n$ over the field of $q^d$ elements, then $Q$ has $m$ (some $m|d$) conjugates $Q_i$ over $\mathbb{F}$ and the product of these conjugates is an irreducible polynomial $P$ in $\mathbb{F}[T]$ so (edit: fixed error pointed out in comments)
$$\sum_i \chi(Q_i)\deg Q_i = (\sum \chi(Q_i))\deg Q = \chi(P)m\deg(Q) = \chi(P)\deg(P).$$
Using this, one checks that the equivalent of $S_n$ over the field extension equals $S_{nd}$ and this gives $\exp (\sum_{n=1}^{\infty} S_{dn} t^n/n )$ is the $L$-function in the extension field, say $L_d(t,\chi)$.
Another way of stating this is
$\prod_{\zeta^d =1} L(\zeta t,\chi)= L_d(t^d,\chi)$.The relation with the zeros follows. (This is e.g. in Weil, Basic Number Theory, Appendix 5, lemma 4 in much more generality and fancier language).
Best Answer
There certainly are L-functions having a functional equation and violating the analogue of RH. (I am not completely sure I understand what type of functional equation you mean by your restriction, but the ones I mention satisfy a functional equation of so to say the usual form, so I hope this is fine.)
Eppstein zeta-functions are a classical example; Davenport and Heilbronn 'On the zeros of certain Dirichlet series' (JLMS) is a classical reference.
For recent investigations in such phenomena see for example Frank Thorne's Analytic properties of Shintani zeta functions.
You might also have a look at the closely related MO question Are there refuted analogues of the Riemann hypothesis? where some information related to this is to be found.
In particular, it seems that for RH one needs Euler product also (not just a functional equation) and then there might be an equivalent; see Frank Thorne's answer on the above mentioned question for more specific information.