The answer to the infinitesimal version is 'no', which makes it very unlikely that the answer to the isometric deformation version is 'yes'. Here is how one can see this:

One can parametrize the upper hemisphere by the unit disk $x^2+y^2\le 1$ conformally by the well-known formula
$$
F(x,y) = \left(\frac{2x}{1+x^2+y^2},\frac{2y}{1+x^2+y^2},\frac{1-x^2-y^2}{1+x^2+y^2}\right).
$$
Let $V(x,y)$ be a vector field along the image of $F$, which can be expressed uniquely in the form
$$
V(x,y) = a(x,y)\,F_x(x,y) + b(x,y)\,F_y(x,y) + c(x,y)\,F(x,y)
$$
for some functions $a$, $b$, and $c$ on the unit disk. Then the condition that $V$ determine an infinitesimal isometric deformation, i.e., $\mathrm{d}F \cdot \mathrm{d}V = 0$, is easily seen to be the system of $3$ equations
$$
a_x-b_y = 0,\qquad a_y+b_x = 0,\qquad c = \frac{2(xa+yb)}{1+x^2+y^2} - a_x\,.
$$
Thus, $a+ib$ must be a holomorphic function of $z = x+iy$, and $c$ is determined in explicitly in terms of $a$ and $b$. In particular,
$$
a + ib = \sum_{n=0}^{\infty} c_n\,z^n
$$
for some complex coefficients $c_n$, $n\ge 0$.

Now, restrict everything to the boundary of the disk, set
$$
E_0(\theta) = F(\cos\theta,\sin\theta),\qquad
E_1(\theta) = F_x(\cos\theta,\sin\theta),\qquad
E_2(\theta) = F_y(\cos\theta,\sin\theta),
$$
so that $E_0$, $E_1$, and $E_2$ are an orthonormal frame field along the boundary curve (i.e., the equator). Let $W(\theta)$ be a vector field along the boundary. It can be written uniquely in the form
$$
W(\theta) = f_0(\theta)\,E_0(\theta)+f_1(\theta)\,E_1(\theta)+f_2(\theta)\,E_2(\theta)
$$
for some $2\pi$-periodic functions of $\theta$. The condition that $W$ furnish an infinitesimal isometric deformation of the boundary curve, i.e., $\mathrm{d}E_0\cdot\mathrm{d}W = 0$, is easily seen to be the differential equation
$$
f_0(\theta) = \frac{\mathrm{d}}{\mathrm{d}\theta}\bigl(\sin\theta\,f_1(\theta)-\cos\theta\,f_2(\theta)\bigr)
$$
Notice that $f_1$ and $f_2$ can be arbitrary $2\pi$-periodic functions of $\theta$. However, if $W$ is to be the boundary value of an isometric deformation vector field $V$ as above, we will have to have
$$
f_1 + i f_2 = a(\cos\theta,\sin\theta) + ib(\cos\theta,\sin\theta)
= \sum_{n=0}^{\infty} c_n\,\mathrm{e}^{in\theta},
$$
i.e., the 'negative' Fourier coefficients of $f_1+if_2$ must all vanish.

Thus, the generic infinitesimal isometric deformation of the boundary cannot be tracked by an infinitesimal isometric deformation of the hemisphere.