The answer is negative as can be seen by putting together these two facts:

 1. There is a bounded convex lower semicontinuous functions defined on a closed and convex subset of $\mathbb R^2$ that is not continuous. One example is [here](https://math.stackexchange.com/questions/2487705/is-a-convex-and-lower-semicontinuous-function-defined-on-a-closed-and-convex-sub). For a large stock of examples, see [this](https://www.jstor.org/stable/23562389).

 2. Every convex, lower semicontinuous and positively homogeneous function on $\mathbb R^n$ is a support function. Theorem 13.2 in Rockefellar's _Convex Analysis_.

For then by (1), let $f$ be a non-continuous bounded convex lower semicontinuous function on a closed subset $D$ of the plane. Extend $f$ to all of the plane by setting it to $+\infty$ outside $D$. This is still convex and lower semicontinuous and it's not continuous on its effective domain. 

Finally, define $g$ on $\mathbb R^3$ as follows:
$$
g(x,y,z) = \begin{cases} 
zf(x/z,y/z) &\text{if }z>0\\
+\infty & \text{if }z\le 0\text{ and }(x,y)\ne 0\\
0 &\text{if }(x,y,z)=0.
\end{cases}
$$
This is convex and lower semicontinuous everywhere and not continuous on the effective domain, and it's positive homogeneous. By (2), $g$ is a support function.