If $\sigma_{H}\ne 0$ and $v_0\ne \theta\ne \sigma^2_{BC}$ then prices are different in BC and Heston model.
Especial case
In the Black-Scholes model the dynamics of $S_t$ under risk neutral measure follow the stochastic process
$$dS_t=(r-q)S_tdt+\sigma_{\color{red}{BC}}S_tdW^{\mathbb{Q}}(t)\tag 1$$
on the other hand
$$dS_t=(r-q)S_t+\sqrt{v_t}S_tdW^{\mathbb{Q}}_1(t)\\
\quad dv_t=\kappa(\theta-v_t)dt+\sigma_{\color{red}{H}}\sqrt{v_t}dW^{\mathbb{Q}}_2(t)\tag 2$$
where $d[W^{\mathbb{Q}}_1(t)\,,\,W^{\mathbb{Q}}_2(t)]=\rho dt$. In CRR model we have
$$\text{Var}[v_t\big{|}v_0]=\frac{v_0\sigma_{\color{red}{H}}^2e^{-\kappa t}}{\kappa}\left(1-e^{-\kappa t}\right)+\frac{\theta\sigma_{\color{red}{H}}^2}{2\kappa}\left(1-e^{-\kappa t}\right)^2\tag 3$$
Now if we set $\sigma_\color{red}{H}=0$ ,then
$$\text{Var}[v_t|v_0] = 0\tag 4$$
This will produce volatility that is time-varying, but $\color{red}{\text{deterministic}}$. Indeed
$$dv_t=\kappa(\theta-v_t)dt\tag 5$$
or
$$v'_t+\kappa v_t=\kappa\theta\tag 6$$
Equation $(6)$ is a linear ordinary differential equation. We can show easily
$$v_t=\theta+c\,e^{-\kappa t}\quad ,\quad c\in\mathbb{R}\tag 7$$
Set $v_0=\theta=\sigma_{\color{red}{BC}}^2$. Therefore $c=0$ and $v=\sigma_{\color{red}{BC}}^2\,.$
$\color{red}{\text{Warning}\,!}$
In the Heston Model we have
\begin{align}
C(t\,,{{S}_{t}},{{v}_{t}},K,T)={{S}_{t}}{{P}_{1}}-K\,{{e}^{-r\tau }}{{P}_{2}}\tag 8
\end{align}
where,for $j=1,2$
\begin{align}
& \mathbb{P}_j({{x}_{t}}\,,\,{{v}_{t}}\,;\,\,{{x}_{T}},\ln K)=\frac{1}{2}+\frac{1}{\pi }\int\limits_{0}^{\infty }{\operatorname{Re}\left( \frac{{{e}^{-i\phi \ln K}}{{f}_{j}}(\phi ;t,x,v)}{i\phi } \right)}\,d\phi \tag 9 \\
& {{f}_{j}}(\phi \,;{{v}_{t}},{{x}_{t}})=\exp [{{C}_{j}}(\tau ,\phi )+{{D}_{j}}(\tau ,\phi ){{v}_{t}}+i\phi {{x}_{t}}]\tag {10} \\
\end{align}
and
\begin{align}
& {{C}_{j}}(\tau ,\phi )=(r-q)i\phi \,\tau +\frac{a}{{{\sigma_{\color{red}{H}} }^{2}}}{{\left( ({{b}_{j}}-\rho \sigma_{\color{red}{H}} i\phi +{{d}_{j}})\,\tau -2\ln \left(\frac{1-{{g}_{j}}{{e}^{{{d}_{j}}\tau }}}{1-{{g}_{j}}}\right) \right)}} \tag{11}\\
& {{D}_{j}}(\tau ,\phi )=\frac{{{b}_{j}}-\rho \sigma_{\color{red}{H}} i\phi +{{d}_{j}}}{{{\sigma_{\color{red}{H}} }^{2}}}\left( \frac{1-{{e}^{{{d}_{j}}\tau }}}{1-{{g}_{j}}{{e}^{{{d}_{j}}\tau }}} \right) \tag{12}\\
\end{align}
such that
\begin{align}
& {{g}_{j}}=\frac{{{b}_{j}}-\rho \sigma_{\color{red}{H}} i\phi +{{d}_{j}}}{{{b}_{j}}-\rho \sigma_{\color{red}{H}} i\phi +{{d}_{j}}} \\
& {{d}_{j}}=\sqrt{{{({{b}_{j}}-\rho \sigma_{\color{red}{H}} i\phi )}^{2}}-{{\sigma_{\color{red}{H}} }^{2}}(2i{{u}_{j}}\phi -{{\phi }^{2}})} \\
& {{u}_{1}}=\frac{1}{2}\,,\,{{u}_{2}}=-\frac{1}{2}\,,\,a=\kappa \theta \,,\,{{b}_{1}}=\kappa +\lambda -\rho \sigma_{\color{red}{H}} \,,\,{{b}_{2}}=\kappa +\lambda \,,\ {{i}^{2}}=-1 \\
\end{align}
We can not simply substitute $\sigma_{\color{red}{H}} = 0$ into the pricing functions,
because that will lead to division by zero in the expressions for $C_j(\tau,\phi)$ and $D_j(\tau,\phi)$.
With $\sigma_{\color{red}{H}}=0$, the
Riccati equation reduces to the ordinary first-order differential equation in the Heston's article (1993)
$$\frac{\partial {{D}_{j}}}{\partial \tau }={{p}_{j}}-{{b}_{j}}{{D}_{j}}\tag {13}$$
where $p_j=u_j i\phi-\frac 12 \phi^2$.The solution of this ODE is
$$D_j(\tau ,\phi )=\frac{(i{{u}_{j}}\phi -\frac{1}{2}{{\phi }^{2}})(1-{{e}^{-{{b}_{j}}\tau }})}{{{b}_{j}}}\tag {14}$$
on other hand , Heston showed
$$\frac{\partial {{C}_{j}}}{\partial \tau }=ri\phi +a{{D}_{j}}\tag{15}$$
substitute $(14)$ in $(15)$ and integrate to obtain
$${{C}_{j}}(\tau ,\phi )\ =ri\phi \tau +\frac{a(i{{u}_{j}}\phi -\frac{1}{2}{{\phi }^{2}})}{{{b}_{j}}}\left( \tau -\frac{1-{{e}^{-{{b}_{j}}\tau }}}{{{b}_{j}}} \right)\tag{16}$$
In the case $j=2$ and $\lambda=0$, we have
$$\begin{align}
& {{D}_{2}}(\tau ,\phi )=-\frac{(i\phi +{{\phi }^{2}})(1-{{e}^{-\kappa \tau }})}{2\kappa } \\
& {{C}_{2}}(\tau ,\phi )\ =ri\phi \tau -\frac{\theta (i\phi +{{\phi }^{2}})}{2}\left( \tau -\frac{1-{{e}^{-\kappa \tau }}}{\kappa } \right) \\
\end{align}
\tag {17}$$
We know
$${{f}_{2}}(\phi ;{{\ln S}_{t}},{{v}_{t}})=\exp\left[i\phi\,{\ln S_t}+{{C}_{2}}(\tau \,,\,\phi )+{{D}_{2}}(\tau \,,\,\phi ){{v}_{t}}\right]$$
let $v_0=\theta=\sigma_{\color{red}{BC}}^2$, thus
$$\color{red}{{{f}_{2}}=\exp\left( i\phi \left[\ln {{S}_{t}}+(r-\frac{1}{2}{{\sigma }_{BC}}^{2})\tau \right]-\frac{1}{2}{{\phi }^{2}}{{\sigma }_{BC}}^{2}\tau \right)=\mathbb{E}\left[\exp\left(i\,\phi\,\ln S_t\right)\right]\tag {18}}$$