2 added 162 characters in body

Here is an argument that resticts the structure of such groups of even order, which expands on ARupinksi's comments above (inspiration came from a nice argument in this paper).
Namely, suppose that $1\neq t$ is a real element in $G$, where $G$ is as in the question. Form the following normal subgroups: $$K = \langle t \rangle^G = \bigcap_{\chi(1)=\chi(t)} \ker \chi \quad \text{and} \quad L = \bigcap_{\chi(t) \neq 0} \ker \chi < K ,$$ where the intersections run over irreducible characters $\chi$ of $G$. Then we have $$t^G = K \setminus L,$$ that is, the elements in $K\setminus L$ are all conjugate in $G$ to $t$. In particular, it turns out that $t$ is rational in the sense that it is conjugate to every generator of $\langle t \rangle$, which is equivalent to every character having rational value at $t$.
Why is this true? Well, column orthogonality yields $\DeclareMathOperator{\Irr}{Irr}$ $$0 = \sum_{\chi \in \Irr G} \chi(t)\chi(1) = |G:K| + \sum_{\chi(t)<\chi(1)}\chi(t)\chi(1).$$ Now let $y\in K$ and suppose that $t$ and $y$ are not conjugate. Then plugging in $y$ instead of $1$ in the last formula, we see that $$\sum_{\chi(t)<\chi(1)} (-\chi(t))\chi(y) = |G:K| =\sum_{\chi(t)<\chi(1)} (-\chi(t))\chi(1) .$$ Since $-\chi(t)>0$ and $|\chi(y)|\leq\chi(1)$, it follows that $y$ is in the kernel of every irr char $\chi$ with $\chi(t)< 0$, that is, $y\in L$. Thus $K\setminus L = t^G$.
(Added later:) If $t$ is an involution, then it follows that $x^t=x^{-1}$ for all $x\in L$, in particular $L$ must be abelian, and elements of $L$ are real.

The "dual" argument (exchanging the roles of characters and conjugacy classes) shows the following: Suppose $1\neq \chi$ is real valued, and let $V= \operatorname{\mathbf{V}}(\chi)$ be the vanishing-off group of $\chi$, generated by all group elements on which $\chi$ is non-zero. Then $$\Irr( G/\ker \chi ) = \{ \chi \} \cup \Irr(G/V).$$ From this it follows easily that $V/\ker\chi$ is a conjugacy class of $G/\ker\chi$ and that the only value of $\chi$ besides $0$ and $\chi(1)$ is $-\chi(1)/(|V/\ker\chi|-1)$. (In particular, any real character is rational.) Groups with such an character have been studied by Zhmud, where more information can be found.
I suppose there is also literature on groups having normal subgroups $L\subset K$ such that $K\setminus L$ is a conjugacy class of $G$. The notion of a Camina pair/group seems to be related (see this paper and papers that refer to it).

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Here is an argument that resticts the structure of such groups of even order, which expands on ARupinksi's comments above (inspiration came from a nice argument in this paper).
Namely, suppose that $1\neq t$ is a real element in $G$, where $G$ is as in the question. Form the following normal subgroups: $$K = \langle t \rangle^G = \bigcap_{\chi(1)=\chi(t)} \ker \chi \quad \text{and} \quad L = \bigcap_{\chi(t) \neq 0} \ker \chi < K ,$$ where the intersections run over irreducible characters $\chi$ of $G$. Then we have $$t^G = K \setminus L,$$ that is, the elements in $K\setminus L$ are all conjugate in $G$ to $t$. In particular, it turns out that $t$ is rational in the sense that it is conjugate to every generator of $\langle t \rangle$, which is equivalent to every character having rational value at $t$.
Why is this true? Well, column orthogonality yields $\DeclareMathOperator{\Irr}{Irr}$ $$0 = \sum_{\chi \in \Irr G} \chi(t)\chi(1) = |G:K| + \sum_{\chi(t)<\chi(1)}\chi(t)\chi(1).$$ Now let $y\in K$ and suppose that $t$ and $y$ are not conjugate. Then plugging in $y$ instead of $1$ in the last formula, we see that $$\sum_{\chi(t)<\chi(1)} (-\chi(t))\chi(y) = |G:K| =\sum_{\chi(t)<\chi(1)} (-\chi(t))\chi(1) .$$ Since $-\chi(t)>0$ and $|\chi(y)|\leq\chi(1)$, it follows that $y$ is in the kernel of every irr char $\chi$ with $\chi(t)< 0$, that is, $y\in L$. Thus $K\setminus L = t^G$.

The "dual" argument (exchanging the roles of characters and conjugacy classes) shows the following: Suppose $1\neq \chi$ is real valued, and let $V= \operatorname{\mathbf{V}}(\chi)$ be the vanishing-off group of $\chi$, generated by all group elements on which $\chi$ is non-zero. Then $$\Irr( G/\ker \chi ) = \{ \chi \} \cup \Irr(G/V).$$ From this it follows easily that $V/\ker\chi$ is a conjugacy class of $G/\ker\chi$ and that the only value of $\chi$ besides $0$ and $\chi(1)$ is $-\chi(1)/(|V/\ker\chi|-1)$. (In particular, any real character is rational.) Groups with such an character have been studied by Zhmud, where more information can be found. I suppose there is also literature on groups having normal subgroups $L\subset K$ such that $K\setminus L$ is a conjugacy class of $G$. The notion of a Camina pair/group seems to be related (see this paper and papers that refer to it).