Yes, this is true, but I don't know a reference, so here's a proof (I think). Let
$$
R(A, x) = \frac{x^T A x}{x^T x}
$$
be the Rayleigh quotient. We know that for a symmetric (in fact Hermitian) matrix $A$ and any vector $x$,
$$
R(A, x) \leq \rho(A)
$$
with equality if and only if $x$ is the Perron vector for $A$.
Now suppose that $H$ is your original graph, and that $G = H \vee K_1$ is the cone, and that $A(H)$, $A(G)$ are their respective adjacency matrices. Then let $\lambda = \rho(H)$ and suppose that $v$ is a vector such that $A(H) v = \lambda v$, and recall that all the entries of $v$ can be taken to be non-negative. Renumbering if necessary we can assume
that
$$
A(G) = \left[
\begin{array}{cc}
A(H)&j\\
j^T&0
\end{array}
\right],
$$
where $j$ is the all-ones vector.
Our aim is to exhibit a vector $w$ such that $R(A(G), w) \geq \lambda+1$ thus proving the inequality. So let $w$ be the vector obtained by adjoining an additional coordinate equal to $1$ to $v$. It then follows that
$$
A(G) w= \left[
\begin{array}{cc}
A(H)&j\\
j^T&0
\end{array}
\right]
\left[
\begin{array}{c}
v\\
1
\end{array}
\right]
=
\left[
\begin{array}{c}
\lambda v + j\\
j^T v
\end{array}
\right]
$$
and
$$
w^T w = v^T v + 1.
$$
Therefore the Rayleigh quotient
$$
R(A(G), w) = \frac{\lambda v^T v + 2 j^T v} { v^T v + 1}.
$$
Now assume that $v$ is normalised so that its maximum entry is $1$ and again recall that every entry of $v$ is non-negative. Then we have
the inequalities $j^T v \geq v^T v$ (because each $0 \leq v_i \leq 1$) and $j^T v \geq (\lambda +1)$ (because if $i$ is the index such that $v_i = 1$ then $(Av)_i = \lambda$ and $j^T v \geq (Av)_i + v_i$). Putting it all together we have
$$
R(A(G), w) \geq \frac{\lambda v^T v + v^T v + (\lambda + 1)}{v^Tv + 1} = \lambda+1.
$$
